Selected topics in Complex Analysis MATH Winter (draft)

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1 Selected topics in Complex Analysis MATH Winter (draft) François Monard March 23,

2 2 Lecture 1-01/05 - Recalls from 427 Algebraic manipulations of complex numbers. See for instance [Taylor, Ch. 1]. Addition, multiplication, conjugation, modulus, inversion of complex numbers. Field properties (C is a field, algebraically closed). Polar forms (z = ρe iθ ), standard form (z = x + iy) and their geometrical meaning. Principal argument: θ = Arg z [0, 2π) such that z = z e iθ. n-th roots of unity and how they for a regular n-gon on the unit circle. Applications to solving equations of the type z n = z 0. Unlike real numbers, there is NO ordering of the complex numbers. Expressions like z 1 < z 2 or z 1 z 2 make no mathematical sense. However, expressions like z 1 z 2 are acceptable (and extremely important!). Triangle inequality and reverse triangle inequality: z w z + w, z, w C (triangle inequality) z w z w, z, w C (reverse triangle inequality). Geometric series: the series n 0 zn converges to 1 1 z on the unit disk D 1(0) = { z < 1}. The exponential function e z = n 0 zn n! and its implications: complex trigonometry, Euler s formula. e iθ = cos θ + i sin θ, θ R (Euler s formula). Complex differentiability: Let U C open, f : U C and z 0 U. We say that f is complexdifferentiable (or differentiable, or analytic, or holomorphic) at z 0 if lim 0 +h) f(z 0 ) f(z h 0 h exists, in which case we call it f (z 0 ). As examples, the exponential function, polynomials, are all differentiable over C. The conjugation map z z is nowhere differentiable, z 1 z is differentiable over C {0}. f is differentiable on U C if and only if u = R(f) and v = Im (f) satisfy the Cauchy-Riemann equations x u = y v and y u = x v throughout U. In particular, u and v are harmonic on U, that is, they satisfy xx u + yy u = xx v + yy v = 0 on U. Path integrals and Cauchy s formula. For f : U C with U open and γ : I = [a, b] U a differentiable curve, we define the path integral γ f dz := b a f(γ(t))γ (t) dt. This integral does not depend on the parameterization of γ, can be extended to piecewise differentiable curves.

3 3 The following ML-estimate holds: γ f dz l(γ) max f(γ(t)), t I where we have defined the length of γ to be l(γ) := b a γ (t) dt (also independent of parameterization). The Jordan curve theorem states that if γ : I C is a simple closed curve (a.k.a. a Jordan curve), then C γ(i) consists of two open connected components, one unbounded Ext(γ) and one bounded Int(γ). Cauchy s integral formula/theorem: If γ is a positively oriented Jordan curve, γ(i) U open and Int (γ) U, then for any f analytic on U, γ f(z) dz = 0. for any f analytic on U and every z 0 Int (γ), 1 f(z) dz = f(z 0 ). 2πi γ z z 0 As a consequence, a function analytic on U admits local power series expansions (PSE): if f : U C and z 0 U, U open, then there exists ρ > 0 such that for every z D ρ (z 0 ), f(z) = k=0 c k (z z 0 ) k, where c k = 1 2πi z = ρ 2 f(z) (z z 0 ) k+1 dz = f (k) (z 0. k! This implies that as soon as f is complex-differentiable on U, it is of class C on U and so is any of its derivatives. Liouville s theorem states that if f is analytic, bounded on C, then f must be a constant. Morera s theorem states the converse of Cauchy s integral formula: if f : U C is continuous and f(z) dz = 0 for any triangle U, then f is analytic on U. For U C open, we say that E U is a discrete subset of U if every point z 0 U has a neighborhood which contains no points of E other than z 0 itself. The identity theorem states that if f, g : U C are analytic on U open and f = g on a non-discrete set of U, then f g. Singularities: Most singularities of analytic functions we encounter are of two natures: jumps across a curve or isolated singularities: Jump across a curve: this is the case for instance of f(z) = Log z = ln z + i arg z where one needs to pick a certain branch for the argument: if arg z is chosen to be in [α, α + π), then f has a jump discontinuity through the half-line {te iα, t > 0}. Functions defined in terms of the log, such as non-integer powers, will inherit some jump discontinuities as well.

4 4 isolated singularities: f has an isolated singularity at z 0 if there exists ρ > 0 such that f is analytic on D ρ (z 0 ) {z 0 } (e.g. z 1 z at z 0 = 0). Isolated singularities divide into 3 cases: Removable singularities: if lim z z0 f(z) exists, or if f is bounded on D ρ (z 0 ) {z 0 }, then f can be given a value at z 0 which makes it analytic there. Pole of order k: there exists k N, g analytic on D ρ (z 0 ) with g(z 0 ) 0 such that f(z) = g(z) (z z 0 ) k on D ρ (z 0 ). This is also the case where lim z z0 f(z) = in the sense that for every M > 0 there exists r > 0 such that z z 0 < r implies f(z) > M. Essential singularities: If lim z z0 f(z) does not exist even as an infinite, z 0 is an essential singularity. Ex: z e 1 z at z 0 = 0. In this case, for every r > 0, f(d r (z 0 ) {z 0 }) = C. We say that f : U C is meromorphic on U iff f is analytic on U except at a discrete set of isolated poles (note that functions with essential singularities are NOT meromorphic). Laurent series: For z 0 C and R > r 0, define an annulus A = A(z 0, r, R) = {z C, r < z z 0 < R}. If f : A C is analytic on A, then f admits a Laurent series expansion f(z) = + k= c k (z z 0 ) k, where c k = 1 f(z) dz, 2πi z z 0 =ρ (z z 0 ) k+1 where ρ (r, R). In this series, negative powers of (z z 0 ) may be due to the singularities of f that lie inside D r (z 0 ). When z 0 U is an isolated sigularity of f : U C (U open), then considering the Laurent series of f on an annulus A(z 0, 0, ρ) with ρ small enough, we define the Residue of f at z 0 to be Res (f, z 0 ) = c 1 = 1 f(z) dz, 0 < r < ρ. 2πi z z 0 =r The Residue theorem states that if U C is open, γ : I U is a Jordan curve with Int (γ) U and f is meromorphic on U with no poles on γ(i), then Int (γ) encloses finitely many isolated singularities of f, call them (z 1,..., z m ), γ f(z) dz = 2πi m k=1 Res (f, z k).

5 5 Lecture 2-01/07 - Counting zeroes and poles. Rouché s theorem. Recall the Residue theorem, which allows to compute curve integrals of meromorphic functions in terms of their residues at the poles enclosed by that curve: Theorem 1. Let f : U C meromorphic on U C open with γ : I U a Jordan curve with Int (γ) U, and such that f has no poles on γ(i). Then (i) Int (γ) encloses finitely many poles of f, call them z 1,..., z m. (ii) γ m f(z) dz = 2πi Res (f, z j ). j=1 We now show how this theorem can be used to compute the number of zeros and poles of f enclosed in a given region of the complex plane. When f is meromorphic, we have the following local representation of f: near a zero or pole z 0 : there exists an integer k 0, ρ > 0 and g analytic on D ρ (z 0 ) (with g(z 0 ) 0) such that f(z) = (z z 0 ) k g(z) for z D ρ (z 0 ). Differentiating this equality, we obtain f (z) = k(z z 0 ) k 1 g(z) + (z z 0 ) k g (z), z z 0 < ρ, so that f f (z) = k z z 0 g(z) + g (z), z z 0 < ρ. at any other point, f f is analytic. As a conclusion, f f is meromorphic wherever f is, and it has simple poles exactly located at the zeroes and poles of f, with residue the multiplicity of that zero/pole (with a negative sign if it is a pole). Combining this with the Residue theorem, we arrive at the next theorem. Theorem 2. Let f : U C meromorphic on U C open. Let γ : I U a Jordan curve with Int(γ) U with no zeros or poles of f on γ(i). Then f (z) ( ) f dz = 2πi Res f(z) f, z j = (2πi)(N 0 N ), γ z j poles of f f where N 0 /N denote the numbers of zeros and poles of f inside γ, counted with multiplicity.

6 6 Example 1. Let f(z) = (z 1)(z i)3. Compute (z 2) 5 γ f(z) dz for the following contours: (a) γ = D 1 ( 1), (b) γ = D 1.5 (0), (c) γ = D 1.5 (2). Theorem 2 allows us to prove Rouché s theorem. Theorem 3 (Rouché s theorem). Let f, g : U C analytic on U C open, γ : I U a Jordan curve with Int (γ) U. Assume f has no zero on γ(i) and f(z) g(z) g(z), z γ(i). (1) Then f and g have the same number of zeros inside γ, counting multiplicities. You may think of f = (f g) + g = small + BIG on γ(i) in the sense of (1): if f g is small enough on γ(i), it does not perturb the number of zeros of g inside γ. Proof of Rouché s theorem. Due to (1), g has no zeros on γ(i) either. Define h := f g, meromorphic on U, with no poles or zeros on γ(i). By Theorem 2, we have h(z) dz = 2πi(N 0 (h) N (h)) = 2πi(N 0 (f) N 0 (g)), γ where the N s refer to numbers of zeros or poles inside γ. Using the equality above, the theorem will be proved if we can show that the left-hand-side is zero, which we now prove. h h To prove that γ dz = 0, a way to rewrite (1) is, upon dividing by g, f(z) g(z) 1 1 z γ(i) h(γ(t)) 1 1, t I, so the closed curve h(γ(t)) is included in D 1 (1) {0}, a convex set where z 1 z is analytic (that h cannot be zero on γ(i) comes from the fact that neither f nor g is vanishes on γ(i), as we stated above). Then Cauchy s theorem implies dz 0 = h γ z = h (γ(t)) h (z) I h(γ(t)) γ (t) dt = γ h(z) dz. Hence the result. Example 2. Let f(z) = 1 + 2z + 7z 2 + 3z 5. Show that f has exactly two roots inside the unit disc. Answer: Apply Rouché s theorem to g(z) = 7z 2 and (f g)(z) = 1 + 2z + 3z 5. We now state a couple of consequences of Rouché s theorem, which may be proved later: Rouché s theorem can provide a short proof of the fundamental theorem of algebra (see Exercise 1 below). Hurwitz s theorem: Let K C compact (or, equivalently, closed and bounded) and let f k : K C an analytic sequence converging uniformly to f on K. Suppose that f has a zero z 0 K of order m > 0. Then there exists ρ > 0 and K N such that for every k K, f k has exactly m zeros on the disk D ρ (z 0 ) and these zeros converge to z 0 as k.

7 7 Exercises for lecture 2: 1. Prove the fundamental theorem of algebra using Rouché s theorem: any polynomial of the form p(z) = n k=0 a kz k with a n 0 has exactly n roots. (Hint: Apply Rouché s theorem with g(z) = a n z n and f(z) = p(z) g(z) on a disk of large enough radius.) 2. Show that 2z 5 +6z 1 has one root in the disk D 1 (0) and four roots in the annulus A(0, 1, 2) = {1 < z < 2}. 3. How many roots does z 9 + z 5 8z 3 + 2z + 1 have between the circles { z = 1} and { z = 2}. 4. Show that if m and n are positive integers, then the polynomial has exactly n zeros inside the unit disk. p(z) = 1 + z + z2 2! + + zm m! + 3zn, 5. For f(z) = e 1 z and f(z) = α z e 1 z where α C is fixed, compute directly z =1 dz. Conclude that in the case where a function has an essential singularity, no analogue of Theorem 2 can be written. f (z) f(z) Material used for this lecture: [Taylor], 4.4, 4.5 [Gamelin], VIII.2

8 8 Lecture 3-01/09 - Consequences of Rouché s theorem: Hurwitz s theorem, counting function. Hurwitz s theorem. Recall that a sequence f n of analytic functions f n : U C converges uniformly to f on K U if and only if ε > 0, n 0, n n 0, f n (z) f(z) < ε, z K. It is a fact (from Math 427) that a uniform limit of analytic functions on K is itself analytic on K. Definition 1. For U C an open domain, a sequence f n : U C of analytic functions converges normally to f on U if it converges uniformly to f on every compact subset of U. For the reason mentioned before this definition, the normal limit f as above is then analytic on U. Beware that normal convergence on U certainly does NOT imply uniform convergence on U (see Ex. 1) below. Examples: 1. The geometric series n 0 zn converges normally on the unit disk D 1 (0). (it s enough to show uniform convergence on every closed disk D r (0) with r < 1). 2. The local power series of an analytic function converges normally to that function on its open disk of convergence. 3. The Laurent series of a function analytic on an annulus A(z 0, r, R) converges normally to that function on the annulus. Theorem 4 (Hurwitz s theorem). Let U C open and f n : U C a sequence of analytic functions on U, converging normally to a limit f on U. If z 0 U is a zero of order m of f, there exists ρ > 0 and n 0 N such that for every n n 0, f n has exactly m zeros inside D ρ (z 0 ), counting multiplicity. Moreover these zeros converge to z 0 as n. Proof. By local representation, there exists ρ > 0 such that f(z) = (z z 0 ) m g(z) for every z D ρ (z 0 ), with g analytic on D ρ (z 0 ) and g(z 0 ) 0. Upon shrinking ρ a bit, since g is continuous, we can assume that g(z) g(z 0) 2 for every z D ρ (z 0 ), so that, for z z 0 = ρ, f(z) = z z 0 m g(z) = ρ m g(z) ρm g(z 0 ) 2 c > 0. Upon shrinking ρ again, we can assume that D ρ U so that by normal convergence applied to the compact set K = D ρ (z 0 ), there exists an integer n 0 such that for n n 0, max z K f n (z) f(z) c. In particular, for every n n 0 and z z 0 = ρ, we have the estimate f n (z) f(z) c f(z).

9 9 By Rouché s theorem, this implies that f n has the same number of zeros as f inside D ρ (z 0 ), counting multiplicity, i.e., m zeros. That these zeros converge to z 0 is clear because we can shrink ρ as much we want, f will still have m zeros inside D ρ (z 0 ), and following the proof above, for n large enough, f n much have its zeros inside D ρ (z 0 ). Example: As a consequence, because the sequence f n (z) = n k=0 zk k! converges normally to e z on C, we can show that for every R > 0 there exists n 0 such that for every n n 0, f n has no zero inside D R (0). This is striking since f n has exactly n roots. This tells us that somehow, all the roots we add at every step should all escape at! The counting function. Definition 2. We say that (i) f attains m times the value w 0 at z 0 iff the function f(z) w 0 has a zero of order m at z = z 0. (ii) f attains m times the value w 0 on U iff the function f(z) w 0 has m zeros on U, counting multiplicity. Examples: The function f(z) = z n is such that f(0) = 0 and f(z) 0 has a zero of order n at z = 0. So it attains n times the value w 0 = 1 at z 0 = 0. Saying that a function attains 5 times a given value w 0 does not necessarily mean that the value w 0 is attained at 5 different places! However this is the correct definition in order to make the following strong statement, which we will prove next class: Theorem: let f : U C analytic and assume that f attains m times the value w 0 at some z 0 U. Then there exist ρ > 0 and δ > 0 such that for every w D δ (w 0 ), f attains m times the value w in D ρ (z 0 ).

10 10 Exercises for lecture 3:: 1. Prove that the sequence f n (z) = n k=0 zk (a) converges uniformly to f(z) = 1 1 z on D r(0) = { z r} for every 0 < r < 1 (you may use the Weierstrass M-test [Taylor, Theorem p77]). (b) does not converge uniformly to f on D 1 (0) = { z < 1}. 2. Let the sequence f n (z) = n k=0 zk k!. (a) Prove that for every R > 0, the sequence f n converges uniformly to f(z) = e z on D R (0). (b) Modify the arguments in Hurwitz s theorem to show that for any R > 0, there exists n 0 such that for every n n 0, f n has no roots inside D R (z 0 ) (Recall that z e z has no roots on C). 3. Let the series f n (z) = z2k+1 k=0 ( 1)k (2k+1)!. Find its pointwise limit and prove, using Hurwitz s theorem, that there exists n 0 such that for every n n 0, f n has exactly one zero inside D π (0) How many times, on C, does the function z attain (a) the value 8? (a) the value 4? 5. Let f(z) = P (z) Q(z) a rational function, where P, Q are polynomials of respective degrees d P d Q, and suppose that the roots of P are all distinct from the roots of Q. Given any value w C, how many times does the function f attain the value w on C? Hint: find that solving f(z) = w amounts to solving a certain polynomial equation. Remark: Using topological facts, 1.(a) is equivalent to saying that f n converges normally to its limit on D 1 (0). Exercise 1 shows how different uniform convergence on U and normal convergence on U are. Material used for this lecture: [Gamelin], VIII.3, VIII.4

11 11 Lecture 4 - Counting function, inverse functions and critical points. In this lecture, we prove that the counting function of an analytic function is locally constant. Depending on that constant value m N, we discuss the local behavior of the corresponding function: if m = 1, then the function is locally invertible; if m > 1, then the function is, near the central point, an m-to-1 map. The counting function. When f is analytic on U open and attains the value m times at z 0, since the function f(z) w 0 is analytic on U, it has a local representation near z 0 of the form f(z) w 0 = (z z 0 ) m g(z) with g analytic near z 0 and g(z 0 ) 0. Since this zero is isolated, we can find ρ > 0 such that f(z) w 0 does not vanish on D ρ (z 0 ), and we denote δ = min z z0 =ρ f(z) w 0. ρ is chosen so that z 0 is the only zero (of order m) inside the disk z z 0 < δ, so that Theorem 2 gives us that 1 f (z) dz = m. 2πi z z 0 =ρ f(z) w 0 What happens then for w near w 0? We would now like to turn the left-hand side into a function of w and see what we can say about it. For w w 0 < δ, we define the following counting function N(w) = 1 f (z) 2πi z z 0 =ρ f(z) w dz, w w 0 < δ. (2) We will prove in the following theorem that this function is well-defined and, in fact, a constant function near w 0. Theorem 5. Let f : U C a nonconstant, analytic function on U open. Suppose f attains m times the value w 0 at z 0 U, then there exist ρ > 0 and δ > 0 such that for every w D δ (w 0 ), f attains m times the value w on D ρ (z 0 ). Proof. Let δ, ρ and the function N(w) defined as in (2). By assumption, N(w 0 ) = m. We are left to show that the function N(w) is well-defined and analytic on D δ (w 0 ). By Theorem 2, N(w) precisely counts the number of zeros of f(z) w inside D ρ (z 0 ), so it s an integer-valued function. Since it is analytic (hence continuous), it must be constant. We now prove that N(w) is well-defined and analytic. We first need to show that N is well-defined, which only requires to show that f(z) w does not vanish on { z z 0 = ρ} for wny w D δ (w 0 ): this is because we have defined δ so that f(z) w 0 δ for z z 0 = ρ and if w w 0 < δ, the reverse triangle inequality implies f(z) w = f(z) w 0 (w w 0 ) f(z) w 0 w w 0 > δ δ = 0, which ensures that (f(z) w) does not vanish on { z z 0 = ρ}: N(w) is well-defined on D δ (w 0 ).

12 12 Second, we show that N is a complex-differentiable function of w on D δ (w 0 ): indeed, it is a parameter-dependent integral, whose integrand is clearly differentiable w.r.t. w with derivative f (z) w f(z) w = f (z). Since the denominator still does not vanish on z z (f(z) w) 2 0 = ρ, the integral dz is well-defined and we can set z z 0 =ρ f (z) (f(z) w) 2 N (w) = 1 f (z) 2πi z z 0 =ρ (f(z) w) 2 dz. This shows that N is differentiable, hence continuous. Since N is integer-valued, it must be constant throughout D δ (w 0 ). So the counting function of an analytic function is locally constant (in the value space). The number of times a function attains a value w 0 at z 0 is related to the order of z 0 as a critical point: Definition 3. z 0 is a critical point of f of order m 1 > 0 (with critical value w 0 = f(z 0 )) iff f (z 0 ) = f (z 0 ) = = f (m 1) (z 0 ) = 0, f (m) (z 0 ) 0. We see that, with the quantities defined above, terms of the form f (j) (z 0 )(z z 0 ) j disappear for 1 j m 1 in the power series expansion of f, so that f can locally be written as f(z) = f(z 0 ) + (z z 0 ) m g(z), g(z 0 ) 0, g analytic near z 0. So z 0 is a critical point of f of order m 1 if and only if f attains m times the value f(z 0 ) at z 0. They mean essentially the same thing, though they describe different phenomena. The remainder of the lecture consists in clarifying the following claims: let f : U C analytic, z 0 U and f(z 0 ) = w 0. (i) if z 0 is non-critical, i.e. f (z 0 ) 0, then f is locally invertible near z 0 and its local inverse is analytic. (ii) if z 0 is a critical point of order m 1, there exists ρ > 0 and δ > 0 such that the restriction f : D ρ (z 0 ) D δ (w 0 ) is a m-to-1 map except at z 0 where it attains m times the value w 0. Local behavior near a non-critical point: Point (i) is summarized in the following theorem: Theorem 6. Let f : U C, z 0 U non-critical (i.e. f (z 0 ) 0), set w 0 = f(z 0 ). Then there exists ρ > 0 and δ > 0 such that for every w D δ (w 0 ), there exists a unique z w D ρ (z 0 ) such that f(z w ) = w. Such a z w is given by the explicit expression z w = f 1 (w) = 1 2πi z z 0 =δ in particular, f 1 is analytic on its domain of definition. zf (z) f(z) w dz,

13 13 Proof. The unique existence of z w follows directly from Theorem 5. For fixed w, we define on D ρ (z 0 ) the mapping h(z) = zf (z) f(z) w. Clearly, h is analytic, its denominator does not vanish on z z 0 = ρ and h only has a simple pole at z = z w (it is simple because the counting function equals 1!), then the residue theorem gives: Res (h, z w ) = 1 h(z) dz. 2πi z z 0 =ρ Moreover, for simple poles, another way to compute the residue is z z w Res (h, z w ) = lim (z z w )h(z) = lim z z w z z w f(z) f(z w ) zf (z) = z w. Hence the result. Since f 1 is expressed as a paremeter-dependent integral with integrand analytic in w, f 1 is itself analytic. Local behavior near a critical point: Point (ii) is now justified more loosely, as we would need to develop more theory to prove some of the statements. For m > 1, let us start with studying the function f(z) = z m near z = 0. Clealy, z 0 = 0 is a critical point of order m 1, with value w 0 = 0. For any w 0, the equation f(z) = w has exactly m distinct solutions, given by z k = w 1 m e map from C {0} to itself. arg w+2kπ i m, for 0 k m 1. So f is essentially an m-to-1 We now briefly explain why this property is still satisfied locally when an analytic function f has a critical point of order m 1 at z 0, with value f(z 0 ) = w 0. Write f near z 0 as f(z) = w 0 + (z z 0 ) m h(z) with h analytic near z 0 ( and h(z 0 ) 0. The latter property allows to define an analytic function h 1 m (z) near z 0 such that h 1 m m (z)) = h(z), so that we may rewrite f as f(z) = w 0 + (g(z)) m, g(z) := (z z 0 )h 1 m (z), z near z0. The function g is analytic near z 0 and g (z 0 ) = h 1 m (z 0 ) 0, so by Theorem 6, g is invertible. Now, for w near w 0, we see that inverting the relation f(z) = w amounts to inverting the relation f(z) w 0 = (g(z)) m, which, when f(z) w 0 0, yields m distinct solutions given by ( z k (w) = g 1 (f(z) w) 1 m e i 2kπ m ), 0 k m 1. The dependency of z k s on w cannot be made analytic on the D δ (w 0 ), but on D δ (w 0 ) minus a slit. These m domains of analyticity can be glued together to form the domain of definition of a single-valued analytic function (it is however defined on an m-sheeted surface). Such constructions pertain to the field of Riemann surfaces. As we will see later, critical points play a fundamental role in several fields, for instance: in the theory of conformal mappings, an analytic function is conformal (i.e. angle-preserving) at z 0 if and only if z 0 is non-critical. complex dynamics, when dealing with questions of connectedness and Julia sets, and they are at the core of the Mandelbrot set s definition.

14 14 Exercises for lecture 4: 1. Using the geometric series, find the power series expansion of the function N(w) on D δ (w 0 ), by writing 1 f(z) w as a power series in (w w 0). 2. How many critical points, counting multiplicity, does a polynomial of degree m have in the complex plane? Justify your answer. 3. Find and plot the critical values of f(z) = z and of its iterates f(f(z)) = (z 2 + 1) and f(f(f(z))). Hint: use the chain rule. 4. Let f(z) be a polynomial of degree m. How many (finite) critical points does the N-fold iterate f f (N times) have? Describe them in terms of the critical points of f(z). Material used for this lecture: [Gamelin] VIII.4, VIII.5. The drawing on p237, which I re-did in class, is a good visualization of what happens near a critical point.

15 15 Lecture 5 - Conformal mappings - Introduction Let us first regard C as R 2 for an introduction. Let I = ( 1, 1) and two curves γ 1, γ 2 : I R 2 intersecting at z 0 = (x 0, y 0 ) = γ 1 (0) = γ 2 (0). Their tangent vectors γ i (0) = d dt γ i(t) t=0 make an angle θ at z 0, that is to say, we have cos θ = γ 1(0) γ 2 (0) γ 1 (0) γ 2 (0), sin θ = γ 1 (0) γ 2(0) γ 1 (0) γ 2 (0), where (v x, v y ) = ( v y, v x ). Let a function h = (u(x, y), v(x, y)) : R 2 R 2 and denote w 0 = h(z 0 ). The image curves Γ 1 = h γ 1 and Γ 2 = [ h γ 2 ] intersect at w 0, and using the chain rule (in two real variables), xu upon denoting J h = yu xv yv the Jacobian matrix of h, the speed vectors at w 0 are given by Γ 1 (0) = J h (z 0 ) γ 1 (0) and Γ 2 (0) = J h (z 0 ) γ 2 (0). Now if h is complex-differentiable, this means that seen as a R 2 R 2 function, its Jacobian matrix assumes the form J h (x, y) = [ ] a b b a by virtue of the Cauchy-Riemann equations. In fact, a, b are such that if z = x + iy, then the complex derivative h (z) equals a + ib. First note by direct inspection that, [ ] [ ] [ ] Jh T a b a b J h = = (a 2 + b ). b a b a 0 1 This implies directly that if a 2 + b 2 0, we have the identities Γ 1 (0) Γ 2 (0) Γ 1 (0) Γ 2 (0) = γ 1(0) γ 2 (0) γ 1 (0) γ 2 (0) and Γ 1 (0) Γ 2 (0) Γ 1 (0) Γ 2 (0) = γ 1 (0) γ 2(0) γ 1 (0) γ 2 (0), so that the angle between ( Γ 1 (0), Γ 2 (0)) and ( γ 1 (0), γ 2 (0)) are the same. In the situation above, we say that h is conformal or angle-preserving at z 0. In general we say that h is conformal on U C iff h is conformal at every point of U. Note that for this property to be satisfied, we needed that h (z 0 ) 0. It turns out that if h (z 0 ) = 0, then h is NOT conformal at z 0 (in fact, if z 0 is a critical point of order m 1, h magnifies angles by a factor m there). Theorem 7. h is conformal at z 0 if and only if h (z 0 ) 0 (i.e. z 0 is non-critical). Example 3. Look at the image curves near z 0 = 0 of γ 1 (t) = t and γ 2 (t) = it for t ( 1, 1) via the mapping z z 2. Definition 4. U, V C are conformally equivalent, denoted U C.E. V or U C.E. V, if there exists a diffeomorphism f : U V, conformal at every point of U. f is called a conformal equivalence. Combining this with Theorem 7, we deduce directly

16 16 Theorem 8. U is conformally equivalent to V if and only if there exists f : U V, complexanalytic such that for every z U, f (z) 0. Note that the inverse function theorem states that if f : U V is analytic with f (z) 0 for every z U, then f is bijective with analytic inverse f 1 : V U. Moreover, we can easily establish that for every w V, (f 1 ) 1 (w) =, so that f : U V is conformal on U if and f (f 1 (w)) only if f 1 : V U is conformal on V. Now we would like to classify which domains are conformally equivalent to which. In order to answer that question, let us first remark that Being conformally equivalent is an equivalence relation: reflexive: U C.E. U. symmetric: if U C.E. V, then V C.E. U. transitive: if U C.E. V and V C.E. W, then U C.E. W. This takes us to the following questions, which we will partially answer over the next few weeks: 1. Given U, V C when can we say that U and V are conformally equivalent? 2. If they are, how do we construct a conformal equivalence? 3. If there is more than one, how many of them are there, and how do we construct all of them? Answering these questions will consist, sometimes, in constructing a library of elementary conformal equivalences, and combining them appropriately by the transitive property described above. Draw pictures when covering the following examples! Example 4. Construct a conformal equivalence from the unit disk U = D 1 (0) to the right-half plane V = {Rz > 0}. Answer: w = f(z) = 1+z 1 z. One can check by direct inspection that f is invertible with inverse z = w 1 1+w and that z < 1 if and only if Rw > 0. Example 5. Construct a conformal equivalence from the unit disk U = D 1 (0) to the top-right quadrant W = {Rz > 0, Im z > 0}. Answer: use V = {Rz > 0} as an intermediate: we can map U to V conformally using f from the previous example, then V from W by finding an appropriate rotation and an appropriate power of z (to tune the aperture of the domain). The map g : V W given by g(z) = iz does the job, so the final answer is h : U W given by h(z) = g(f(z)) = i 1+z 1 z. Example 6. Let f(z) = log z = ln z + i arg (0,2π) (z). If arg z (0, π) and z R >0, w = log z describes the strip V = {Rw R, Im w (0, π)} and f never vanishes on U. Therefore, f is a conformal equivalence from the upper-half-plane to the strip V. In particular, w e w is the inverse conformal equivalence.

17 17 Exercises for lecture 5: In problems 2 through 5, think about constructing equivalences out of simpler ones either from previous problems or examples from the lecture. 1. Check that being conformally equivalent is an equivalence relation. 2. Find a conformal equivalence from the half-plane {z : Rz > 1} to the unit disc D 1 (0). 3. Find a conformal equivalence from D 1 (0) {0} = A(0, 0, 1) to C D 1 (0) = {z : z > 1}. 4. Find a conformal equivalence from the top-right quadrant {z : arg z (0, π 2 )} to C (, 0]. 5. Find a conformal equivalence from the upper half-plane to C (, 1]. Material used for this lecture: [Taylor], 6.1

18 18 Lecture 6 - The Riemann Sphere The purpose of this lecture is to compactify C by adjoining to it a point at infinity, and to extend to concept of analyticity there. Let us first define: a neighborhood of infinity U is the complement of a closed, bounded set. A basis of neighborhoods is given by complements of closed disks of the form U z0,ρ = C D ρ (z 0 ) = { z z 0 > ρ}, z 0 C, ρ > 0. Definition 5. For U a nbhd of, the function f : U C has a limit at infinity iff there exists L C such that for every ε > 0, there exists R > 0 such that for any z > R, we have f(z) L < ε. We write lim z f(z) = L. Equivalently, lim z f(z) = L if and only if lim z 0 f ( 1 z ) = L. With this concept, the algebraic limit rules hold in the same way that they hold at finite points when limits are finite. Example 7. lim z 1 z = 0. lim z z 2 +1 (z 1)(3z+7) = 1 3. lim z e z does not exist (this is because e 1 z to prove this is that both sequences z n = 1 2nπi and z 1 n = 2πi(n+1/2) sequences e 1 zn and e 1 z n has an essential singularity at z = 0). A way converge to zero, while the converge to different limits, 1 and 0 respectively. Definition 6. lim z z0 f(z) = if for every M > 0, there exists ρ > 0 such that z z 0 < ρ implies f(z) > M. Equivalently, lim z z0 f(z) = if and only if lim z z0 1 f(z) = 0. Combining both definition above, we can say that lim z f(z) = iff for every M > 0, there exists R > 0 such that z > R implies f(z) > M. Example 8. If f has a pole of order k > 0 at z 0, f may be written as f(z) = g(z) (z z 0 ) k g(z 0 ) 0 and g analytic near z 0. Then 1. f(z) = (z z 0) k For f any nonconstant polynomial, lim z f(z) =. with 1 g(z) so lim z z0 f(z) = 0, i.e., lim z z 0 f(z) = Definition 7. The extended complex plane (a.k.a. Riemann sphere) is the topological space C = C { }, where open sets are generated either by open disks D ρ (z 0 ) or neighborhoods of infinity. is now an abstract point which is no different than the other ones. We may study the behavior of a function f : C C at by studying the function f( 1 z ) near z = 0. The question now is: when can such a function be extended as an analytic function f : C C?

19 19 Definition 8. f : C C is analytic at iff the function w f( 1 w ) has a limit in C at w = 0 and (i) if this limit is finite, f( 1 w ) is analytic at w = 0, 1 (ii) if this limit is, is analytic at w = 0. f( 1 ) w Example 9. Show that f(z) = 1 z 1+z is analytic at upon extending it with the value 1 there. Definition 9. f : C C has a pole of order n at if f( 1 w ) has a pole of order n at w = 0 (we allow n = 0, which covers the case of a finite limit). Another way of saying that f has a pole at is to say that lim z f(z) exists in C (i.e. either as a finite number, or as ). Theorem 9. f can be extended into an analytic function f : C C if and only if f is meromorphic on C and lim z f(z) exists, either as a complex number or. This extension is done as follows : when f is a meromorphic function on C such that lim z f(z) exists in C, we define f : C C : At z C where f is analytic, define f(z) = f(z). At z C where f has a pole, define f(z) =. At, set f( ) = lim z f(z). Analytic functions on C define conformal maps near every point where f 0. Theorem 10. The map s : C C defined by s(z) = 1 z conformal equivalence of C onto itself. with s(0) = and s( ) = 0 is a Proof. Check that s is one to one and onto: this is because s s = Id. s is analytic at every z 0,. At z =, s( 1 w ) = w is analytic at w = 0 so s is analytic at z =. At z = 0, 1 s(z) = z is analytic at z = 0 so s is analytic at z = 0. Why the Riemann sphere? Consider the sphere in S = {x x2 2 + (x 3 1) 2 = 1} R 3 centered at (0, 0, 1), with North Pole P = (0, 0, 2). S is sitting on the plane {x 3 = 0} C, and we can define the stereographic projection map, for M S {P } to be the unique intersection point between the line P M and the plane {x 3 = 0}. This map is a homeomorphism: it is bijective, continuous with continuous inverse ρ 1. As the neighborhoods of we defined earlier are mapped via ρ 1 to neighborhoods of the North Pole P, we may extend ρ : S {P } C into a homeomorphism ρ : S C by declaring ρ(p ) = and ρ(m) = ρ(m) for any M S {P }.

20 20 In that sense, we can assimilate C as a two-dimensional sphere. We will get other (visual) opportunities to see why in the future. Anticipating a bit, an important property of C is that it is now a compact topological space, unlike C. In particular, every sequence in C has at least one cluster point. This will imply that discrete subsets of C can only be finite, with consequences on analytic functions C C (e.g. the identity theorem).

21 21 Exercises for lecture 6: 1. Show that the function f(z) = 1. z 2 +5 has a limit at, find it, and show that f is analytic at 2. Show that the function f(z) = 2z 1 is not analytic at as a complex-valued function, but it is analytic at as a C -valued function. 3. If f(z) = z4 z 2 1 is considered as an analytic function from C to C, where are the poles of f and what are their orders? 4. Can we consider sin z as an analytic function from C to C? Justify your answer. 5. Prove that if a function f is analytic on C and lim z f(z) = l C exists and is finite, then f is uniformly bounded on C. (By Liouville s theorem, this implies that f is constant. As a consequence, on C, Liouville s theorem can be reformulated as follows: if f is analytic on C and f is finite everywhere (i.e. f(z) for every z C ), then f is constant.) 6. (a) Show that if f has a pole of order k at, then z k f(z) is bounded on some neighbourhood of. (b) Show that if f is an analytic function from C to C which has no poles on C, then f is a polynomial. [Hint: use the previous question and find a way to apply Liouville s theorem as in the proof of Theorem 12 ( = )] (c) Show that an analytic function f : C C can only have finitely many poles in C. [Hint: f( 1 w ) and f(w) are both meromorphic on D 1(0), and you can use the fact that discrete subsets of D 1 (0) can only be finite.] (d) Show that an analytic function f : C C must be a rational function of the form f(z) = P (z) Q(z), with P, Q two polynomials. [Hint: let z 1,..., z d the finite poles of f with multiplicities m 1,..., m d and let Q(z) = d j=1 (z z j) m j. Explain why Q(z)f(z) has no finite poles, has a pole at, and use part (b).] Material used for this lecture: [Taylor], 6.2

22 22 Lecture 7/8 - Linear Fractional Transformations (a.k.a. Möbius transformations) Definition 10. A Linear Fractional Transformation (or Möbius transformation) is a transformation of the form h(z) = az+b cz+d, with a, b, c, d C such that ad bc 0. Example 10. Affine transformations L(z) = az + b (c = 0, d = 1) requires a 0. s(z) = 1 z, i.e. (a, b, c, d) = (0, 1, 1, 0). Note that for every λ 0, the transformations h a,b,c,d and h λa,λb,λc,λd define the same transformation. One could add the normalizing condition ad bc = 1, though it is sometimes more readable to not normalize. Let us first notice that every LFT can be written as either (i) h = L 1 for L 1 some affine transformation, or (ii) as h = L 1 s L 2 for L 1, L 2 two affine transformations. Indeed, if c = 0, then (i) is satisfied, and if c 0, we write h(z) = az + b a ad cz + d = c (cz + d) + b c cz + d = a c + bc ad c 1 cz + d, so that, upon defining L 1 (z) = bc ad c z + a c and L 2(z) = cz + d, then h = L 1 s L 2. Definition 11. For U C an open set, h is a conformal automorphism of U (i.e. h Aut (U)) iff h : U U is conformal and bijective. Proposition 11. For U C an open set, Aut (U) is a group under composition. Proof. The composition of conformal, bijective maps is clearly conformal and bijective. Composition is always associative ((f g) h = f (g h)), the identity element is Id : U U and if f Aut (U), then so it f 1. Theorem 12 (Characterization of Aut (C )). h Aut (C ) if and only if h is a linear fractional transformation. This also suggests that the Möbius transformations form a group, as we will see later. Proof. ( = ) Suppose h is an LFT. h is bijective: indeed, if h is an affine map L(z) = az + b, then it is bijective with inverse L 1 (w) = 1 a (w b); if h is of the form L 1 s L 2, i.e. a coumpound of three invertible transformations, then h is bijective with inverse h 1 = L 1 2 s L 1 1. Moreover, h is conformal: if h(z) = L(z) = az + b, then h (z) = a 0 for every z C so h is conformal; if h = L 2 s L 1, h is conformal as a composition of conformal functions. ( = ) Suppose f Aut (C ) (so f is one to one, onto and conformal). First case: if f( ) =, then q(z) = s f s Aut (C ) is such that q(0) = 0. But q (0) 0

23 23 since q is conformal, so f has a pole of order 1 at. Since f is one to one on C, has no other preimage by f, so f has no pole on C, i.e. the function z 1 (f(z) f(0)) is analytic and finite everywhere on C : by Liouville s theorem, it is constant on C, then there exists c C such that z 1 (f(z) f(0)) = c, i.e. f(z) = cz + f(0). Second case: If f( ) = k. Let p(z) = 1 z k an LFT so p Aut (C ). By the group property, p f Aut (C ) and p f(z) = 1 f(z) k a, b such that az + b = p f(z) = 1 f(z) k is such that p f( ) =. By the first case, there exists 1 so that f(z) = az+b + k is an LFT. Group property lecture 8 ( ) a b Each non-singular 2 2 complex matrix A = determines a linear fractional transformation φ A (z) = az+b cz+d. A straighforward calculation shows that φ A φ B = φ AB and that φ 1 A = φ A 1, c d where AB is matrix multiplication and A 1 = 1 ( d b ) ad bc c a. In other words, the mapping A φa is a group homomorphism. Most importantly, this property makes it easier to compute the composition of two LFT s on the fly, by computing the product of the corresponding matrices. Lines and circles Definition 12. A circle in C is either a circle in C or l { }, where l is a line in C. Remark: In the sphere model of C, the stereographic projection sends circles in the two-sphere to circles in C, except when the initial circle passes through the North Pole, in which case the image is a line. One can show that the C is a circle in C if and only if there exist a, b reals and w C such that C = C(a, b, w) = {z : a z 2 + wz + w z + b = 0}. Theorem 13. Each linear fractional transformation maps circles in C to circles in C. Proof. Since LFT s are always compounds of affine functions and the inversion s, it suffices to show it for these elementary cases. The case of affine transformations is obvious, as they are rotations and translations. In the case of s, we can establish that s C(a, b, w) if and only if 1 z C(b, a, w), hence, s(c(a, b, w)) = C(b, a, w). The proof is complete. Example 11. What is the image of z = 1 under the transformation h(z) = 2z z i? Answer: the image should be a circle in C which we know is characterized by three point. Since h(i) =, the image has to be of the form l where we compute two points passing through l, for instance, h(1) = 1 + i and h( 1) = 1 i. Therefore, h({ z = 1}) = {Rz = 1} { }. We finally show that a given LFT is completely characterized by its images of three distinct points. Note first that it is straightforward to find, for a triple of distinct points (w 1, w 2, w 3 ) an LFT mapping (w 1, w 2, w 3 ) to (0, 1, ): the transformation h(z) = (w 2 w 3 ) (z w 1 ) does the trick. (w 2 w 1 ) (z w 3 )

24 24 Theorem 14. Given (w 1, w 2, w 3 ) and (z 1, z 2, z 3 ) two triples of distinct points, there exists a unique Linear Fractional Transformation h such that h(w j ) = z j for j = 1, 2, 3. Proof. Step 1: If (w 1, w 2, w 3 ) = (z 1, z 2, z 3 ) = (0, 1, ), then h(z) = z. Indeed, looking for h in the form h(z) = az+b cz+d, we have that h(0) = 0 implies b = 0, h( ) = implies c = 0, then h(z) = a d z and h(1) = 1 implies a d = 1. Step 2: For general triples (w 1, w 2, w 3 ) and (z 1, z 2, z 3 ), define h 1 an LFT mapping (w 1, w 2, w 3 ) to (0, 1, ) and h 2 an LFT mapping (z 1, z 2, z 3 ) to (0, 1, ). Clearly, the transformation h 1 2 h 1 is an LFT and does the trick. If f is another LFT mapping (w 1, w 2, w 3 ) to (z 1, z 2, z 3 ), then h 1 1 f h 2 is an LFT mapping (0, 1, ) to itself, which by step 1 implies h 2 f h 1 1 = Id, which implies f = h 1 2 h 1. Example 12. Find a LFT mapping (1, i, 0) to ( i, 0, 2). Let us first compute a LFT mapping (1, i, 0) to (0, 1, ), it is given by h 1 (z) = i z 1 i 1 z, then an LFT mapping ( i, 0, 2) to (0, 1, ), it is given by h 2 (z) = 2i z+i z 2. The final answer is given by h = h 1 2 h 1, which upon using matrix algebra as above, yields the answer: h(z) = (2 + 2i)z + (2 2i) ( 1 + 5i)z + (1 i). Another way to do it is to notice that i is a zero so the numerator should be proportional to (z i). Identifying the other unknown coefficients, then, should become simpler.

25 25 Exercises for lectures 7/8: 1. Find an LFT that takes the line {Rz = 1} to the circle of radius one centered at What is the image of the unit disk under the LFT h(z) = 2iz z 1? 3. What is the image of the disk of radius 1, centered at 1 + i, under the LFT h(z) = 1+z 1 z? 4. In class, we talked about how an equation of the form a z 2 + wz + w z + b = 0 with a, b reals and w complex is a general equation for a circle in C. Fixing (a, b, w), and using z = x + iy, show how this equation can be recast as either an equation of a circle (x x 0 ) 2 + (y y 0 ) 2 = r 2. In this case, express (x 0, y 0, r) in terms of (a, b, w). or an equation of a line αx + βy = γ. In this case, express α, β, γ in terms of (a, b, w). 5. Compute explicitly the LFT s determined by the following correspondence of triples: (a) (1 + i, 2, 0) (0,, i 1) (b) (0, 1, ) (1, 1 + i, 2) (c) (1, i, 1) (1, 0, 1) Material used for this lecture: [Taylor], 6.3 [MTR] (video)

26 26 Lecture 9 - Classification of Möbius transformations In lecture 7, we saw that LFTs could be represented either as an affine transformation L(z) = az + b or as L 1 s L 2, with L 1, L 2 affine. We now turn on to a second classification, which is motivated by dynamical systems. We say that a set S C is stable under M if M(S) S. Stable curves help us visualize the action of certain maps on the complex plane. This is the kind of approach we will use to understand the action of Möbius transformations. As first examples, stable curves for the transformations below are given by: Example If M(z) = e iθ z for some θ R, any circle of center 0 is stable under M. 2. If M(z) = ρz for some ρ (0, + ), any half-line {te iθ, t (0, + )} is stable under M. 3. If M(z) = ρe iθ z, spirals of equation {z(t) = z 0 ρ t e itθ, t R} are stable under M. 4. If M(z) = z + b, straight lines of equation {z(t) = z 0 + bt, t R} are stable under M. Figure 1: Stable curves for the examples 13, l. to r., 1, 2, 3, 4. In example 13, the first three examples have fixed points at 0 and whereas the last has only one fixed point at. The study of the general case starts with, given a transformation M(z) = az+b cz+d, the computation of its fixed points, solving the equation az + b = cz 2 + dz, which, unless M(z) = z, has at most two distinct solutions. Here and below, we now assume M(z) to be normalized, that is, such that ad bc = 1. Direct resolution of the equation above leads to the following fixed points ξ ± = a d± (d a) 2 +4bc 2c, which upon using ad bc = 1, yields ξ ± = a d + (a + d) c

27 27 We see that M has only one fixed point if and only if a + d = ±2, where ξ + = ξ = a d 2c. This case will be studied last. Case of two fixed points: Suppose M has two distinct fixed points ξ ±. Introduce the Möbius transformation T (z) = z ξ z ξ + mapping ξ to 0 and ξ + to. Then the transformation M = T 1 M T maps 0 to 0 and to. As M is a Möbius transformation fixing 0 and, it can only be of the form M(z) = mz for some m C. We call m the multiplier of the transformation M. We now use the following idea to draw the stable curves of M: since M T = T M, we can derive the following principle that S is stable by M if and only if T 1 (S) is stable by M. Using the fact that when m = e iθ or m = ρ, it is easy to know the stable curves of M, we can then map such curves into stable curves of M by T 1. Direct computation of the transformation M = T 1 M T shows that this multiplier is expressed in terms of the quantity a + d via the relation m = w + w 2 4 w, where w = a + d. (3) w 2 4 Remark 1. The multiplier depends on which fixed point we send to 0 and which one we send to. In our convention, we mapped ξ 0 and ξ +. If we interchanged them by using the change of variable T 1 (z) = z ξ + z ξ instead, then the value of the multiplier obtained would be 1 m, where m is defined in (3). Elliptic transformations. If w R and w 2 < 4, then w 2 4 = iy for y = 4 w 2 R, and thus m = w+iy w iy has modulus one, i.e. there exists θ such that m = eiθ. In this case, M(z) = e iθ z is a rotation and its stable curves are circles centered at 0. Mapping them back via T 1 into the stable curves of M, we obtain the left-hand picture of Fig. 2. Hyperbolic transformations. If w R while w 2 > 4, then m R and m > 0. Therefore, M is a dilation (or contraction) and its stable curves are radial rays. Mapping them back via T 1 into the stable curves of M, we obtain the third picture of Fig. 2. Loxodromic transformations. If w / R, then m = ρe iθ, in which case, the stable curves of M are spirals. Mapping them back via T 1 into the stable curves of M, we obtain the third picture of Fig. 2. Case of a single fixed point: Call ξ = ξ ± = a d 1 2c. Setting T (z) = z ξ so that T 1 (z) = ξz+1 z, we define M = T M T 1. Direct computation of the corresponding transformation yields M(z) = z + 2c a + d, so M is a translation. Due to the relation M T = T M, the stable curves of M are obtained by applying T 1 to the stable curves of M, which are straight lines of direction 2c a+d. With the previous considerations, we have proved the following theorem:

28 28 Figure 2: Stable curves of elliptic, hyperbolic and loxodromic transformations (l. to r.). In this example, ξ = 3 and ξ = 2 + i. Figure 3: Stable curves of a parabolic transformation with double fixed point at ξ = 1 + i.

29 29 Theorem 15. Let M(z) = az+b cz+d a normalized Möbius transformation (i.e. ad bc = 1). If a+d / {±2}, there exists m C and a Möbius transformation T (z) such that T M T 1 (z) = mz. In this case, M(z) is elliptic if a + d R and a + d < 2. hyperbolic if a + d R and a + d > 2. loxodromic if a + d / R. If a+d = ±2, there exists c C and a Möbius transformation T (z) such that T M T 1 (z) = z+c. In this case, M(z) is called parabolic. Up to the bijective change of variable T on the Riemann sphere, one may think of elliptic LFTs as rotations, hyperbolic LFTs as dilations, loxodromic LFTs as a combination of the first two cases, and parabolic LFTs as translations. Remark 2. If the transformation is not normalized (i.e. if ad bc 1), we can normalize it by dividing every coefficient by ad bc. Therefore, when the transformation is not normalized, one may draw the same conclusion as Theorem 15 about the type of transformation, upon looking at the a+d quantity ad bc instead of a + d. Example 14. Let M(z) = 2z+3 3z+5. Compute the fixed points, find its multiplier, its type and draw a rough sketch of its phase portrait. Answer: The transformation is already normalized and we have a + d = 7 so M is hyperbolic with multiplier m = > 0. The fixed points solve the equation 2z + 3 = 3z2 + 5z, so that ξ ± = 1± 5 2. Since m > 1, trajectories leave from ξ and reach toward ξ +. Example 15. Same thing with M(z) = 2z+3 z 1. Answer: It is normalized, so w = a + d = 2 1 = 1 so M is elliptic. Its multiplier is m = = 1+i 3 1 i = 1+i = e 2iπ 3. The fixed points solve the equation z 2 + 3z + 3 = 0, so that ξ ± = 3±i 3 2. Upon defining T (z) = z ξ z ξ +, the transformation M = T M T 1 is a rotation of angle 2π 3. (in particular M M M = Id, so that M M M = Id). Example 16. Same thing with M(z) = 4z + 3 = 4z+3 0z+1. a+d ad bc = Answer: M is not normalized so we compute fixed point is and the second solves 4z + 3 = z, i.e. z = 1. In the case of two finite fixed points, we used to send one to 0, the other to. Here, one is already at so all we need is to send the other one to 0. We do this by defining T (z) = z + 1 and M = T M T 1. Since T 1 (z) = z 1, we quickly arrive at = 5 2, so M is hyperbolic. The first M(z) = T (M(T 1 (z))) = M(z 1) + 1 = 4(z 1) = 4z. So M is a dilation with multiplier m = 4.

30 30 Exercises for lecture 9: 1. In the case where M has two distinct fixed points, write down an expression for M = T 1 M T and deduce (3). The calculation may be sped up by using 2 2 matrix multiplications as in lecture For the following LFTs, compute the fixed points, find the multiplier, the type of the LFT, and draw a rough sketch of its phase portrait (i.e. stable curves). (a) M(z) = 3z + 5. (b) M(z) = z+1 z 1. (c) M(z) = 1 z i. Material used for this lecture: [Needham], 3.VII

31 31 Lecture 10 - Conformal equivalences and Dirichlet problems On a simply connected 1 domain Ω R 2, we would like to solve the following equation for a real-valued function u : Ω R u = 2 u x u y 2 = 0, (on Ω), u Ω = g (prescribed). (4) Such a problem is motivated in abundance by problems in physics: 1. In problems of heat conduction for instance, one may be searching for steady-state solutions of a heat conduction problem u t = k u with initial condition u(x, y, 0) = T 0(x, y) and timeindependent boundary condition u(x, y, t) = g(x, y) for all (x, y) Ω. The temperature distribution, as time becomes large enough, should relax toward a time-independent state u (x, y), which satisfies Laplace s equation u = 0 inside Ω, with boundary condition u Ω = g. 2. In electrostatics, Gauss s law for the electrostatic field states that E = ρ ε 0, with ρ the charge density and ε 0 the electric permittivity of the medium. The electric field is assumed to derive from an eletrostatic potential V (x, y) such that E = V. In the absence of charge ρ 0, the potential satisfies the Laplace s equation: 0 = ρ ε 0 = E = V = V, which, again, may be augmented with prescription of a condition at the boundary V Ω = g. 3. Other applications to, e.g., fluid flow may be found in [Gamelin], [Taylor],... The theory of complex-differentiable functions helps us to find solutions to Laplace s equation because any complex-differentiable function f = u + iv is such that u and v are harmonic. The question is, how can we find a harmonic function fulfilling the prescribed boundary condition, and how do these solutions depend on the domain Ω? We start by stating a couple of properties of equation (4): Linearity principle: If u solves (4) with boundary condition g, and v solves (4) with boundary condition h, and α, β are fixed complex numbers, then αu + βv solves (4) with boundary condition αg + βh. This is mainly because (αu + βv) = α u + β v. Uniqueness principle: If a solution to (4) exists, then is it unique. Indeed, if u 1 and u 2 both solve (4) with the same boundary condition g, then by the linearity principle, u 1 u 2 solves (4) with boundary condition g g = 0. By [Taylor, Theorem 3.5.8] 2, u 1 u 2 must achieve its max and min at points of Ω, where it is identically zero. So u 1 u 2 0 inside the domain, i.e. u 1 = u 2. 1 i.e., Ω has no holes, though we will make this definition more precise later 2 If u : U R is harmonic on an open, connected set U, then u achieves its maximum and minimum values at points of U.

32 32 Now we will be following the agenda below: 1. If the domain is the unit disc D = D 1 (0), we can either call out the solution if the boundary condition calls for an obvious enough solution, or we can appeal to the Poisson integral formula, a formula the gives the solution u in the form of an integral of its boundary values {g(e it ), t [0, 2π)}. 2. Let U h V two conformally equivalent domains via h. Suppose we know how to solve (4) on V. Then the solution of u = 0 (U), u U = g, is given by u = v h, where v is the solution to v = 0 (V ), v V = g h 1. Special cases will include the case where V = D. 3. In some other cases, we will call out certain solutions by inspection and deduce a library of examples from which to draw from. As a first example in building our library: Example 17. Let Ω the pie-slice domain Ω = {z : 0 < Arg z < π 4 }. Find a harmonic function u on Ω such that u = 0 on {Arg z = 0} and u = 1 on {Arg z = π 4 }. Answer: A harmonic function that is constant along rays emanatic from zero is the (principal) argument function Arg z. It is harmonic as the imaginary part of Log z, a complex-differentiable function on Ω. It is zero on {Arg z = 0} but not equal to 1 on {Arg z = π 4 }, though by linearity, we can rescale it by dividing by π 4 and still have a harmonic function. The final answer is u(x, y) = 4 π Arg (x + iy) = 4 π tan 1 y x.

33 33 Exercises for lecture 10: 1. Material used for this lecture: [Taylor],

34 34 Lecture 11 - Solving Laplace s equation on the unit disk D We now focus on solving the following problem: given g (continuous at first, but sometimes not!) defined on the unit circle, find the 3 function u harmonic on D such that u D = g. Case 1: When g is a trigonometric polynomial. Example 18. Find u harmonic on D with u D (e iθ ) = g(e iθ ). 1. g(θ) = cos θ. Answer: Notice that cos θ = R(e iθ ) is the real part, on D, of the function f(z) = z, analytic on D. So u(x, y) = R(f(x + iy)) = x is harmonic and fulfills the boundary condition. 2. g(θ) = cos(2θ). Answer: Notice that cos(2θ) = R(e 2iθ ) is the real part, on D, of the function f(z) = z 2, analytic on D. So u(x, y) = R(f(x + iy)) = R(x + iy) 2 = x 2 y 2 is harmonic and fulfills the boundary condition. 3. g(θ) = cos(nθ) for some n N. Following the same approach, we arrive at the solution u(x, y) = R(z n ) = R((x + iy) n ). 4. g(θ) = cos 2 θ. First realize using trig identities that cos 2 θ = 1+cos(2θ) 2. So using part 2 of the example and the linearity (superposition) principle, we obtain u(x, y) = 1 2 (1 + x2 y 2 ). As a general principle, we realize that if g(e iθ ) = a 0 + n k=1 (a k cos(kθ) + b k sin(kθ)) for some real-valued coefficients a k s and b k s, then g is the real part, on D of the analytic function f(z) = a 0 + n k=1 (a k ib k )z k, so that the general solution is given by ( n ) u(x, y) = a 0 + R (a k ib k )z k. Remark 3 (On Fourier series). In fact, this method would generalize even to infinite sine and cosine series, as long as the series k=0 (a k) 2 + (b k ) 2 is convergent. Moreover, even if g is not expressed as a trigonometric polynomial, if it is continuously differentiable, it can be represented pointwise by a Fourier series whose coefficients are given by a 0 = 1 2π π 0 g(θ) dθ, and for n 1, a n = 1 2π 2π 0 g(θ) cos(nθ) dθ and b n = 1 2π 2π 0 g(θ) sin(nθ) dθ. Then the method to construct u as above would still apply. Case 2: Poisson s integral formula. For any continuous function g on the unit circle, this approach just gives an explicit expression for a solution u in terms of g, via the expression u(re iθ ) = 1 2π 2π 0 k=1 g(e it )P r (θ t) dt, where P r (θ) := A handwavy justification for this formula is the following: 3 Recall that the uniqueness principle says that it is unique 1 r 2 1 2r cos θ + r 2. (5)

35 35 First notice that the function h(z) = 1+z 1 z is analytic on the unit disk (in fact, on C {1}), so its real part u(z) = 1 z 2 is harmonic on D. Moreover, u = 0 on D except at z = 1 where 1 z 2 it is undefined. u can be seen as the harmonic function generated by a point mass at 1. In polar coordinates, u(re iθ 1 r ) = 2 = P 1 2r cos θ+r 2 r (θ). Since is a rotation-invariant operator, we expect that the harmonic solution generated by a point mass at e it is just a rotated version of P r (θ), that is, the function z = re iθ P r (θ t). As a function, g(e iθ ) can be regarded as a weighted superposition of point masses g(e iθ ) = 1 2π 2π 0 g(e it )(2πδ(θ t)) dt, and we know how to extend each point mass harmonically inside D by P r (θ t), so that by linearity principle, the harmonic function u inside D with boundary values g can be written as (5). It is shown in Taylor that the solution u defined in (5) solves the problem we want, namely: if g is continuous, then (i) the function u defined in (5) is harmonic inside D. ([Taylor, Theorem 6.5.3]) (ii) the function u is continuous on D and for every t [0, 2π), lim z e it,z D u(z) = g(e it ). This formula is mostly useful to use on a computer and you should probably not use it to compute solutions by hand, though it is interesting on its own right, as an explicit formula. Remark 4. The previous methods seem to work only if g is continuous. It turns out that such conditions can be weakened, though the solution is then defined in a weaker way as well. Case 3: Using conformal equivalences. Using other methods that merely require knowing a library of examples, we can even solve explicitely some problems where the boundary condition is discontinuous. The strategy we will follow is that from point 2 in the agenda from lecture 10. Example 19. Find u harmonic on D such that u = 2 on the upper-half unit circle and u = 0 on the lower-half unit circle. Answer: The map h(z) = 1+z 1 z maps D to the right-half plane, maps the upper-half unit circle to the ray Γ 1 = {it, t (0, + )}, and the lower-half circle to the ray Γ 2 = { it, t (0, + )}. Now, a harmonic solution v of v = 0 (R(z) > 0), v Γ1 = 2, v Γ2 = 0, is given by, up to translation and scaling, the argument function (this is because the argument function is harmonic and constant along rays through zero). Upon adjusting the values properly, we find v(z) = 2 ( arg π ( π,π) (z) + π ) = π arg ( π,π)(z). Then the solution we seek is u(z) = v(h(z)): since v is harmonic, it is the real part of an analytic function f and since h is conformal, f h is analytic, so u = R(f h) is harmonic. Moreover, u satisfies the appropriate boundary condition. Conclusion, u(x, y) = π arg 1 + z ( π,π) 1 z = y π tan 1 1 x 2 y 2.

36 36 Exercises for lecture 11: 1. Let y 0 < y 1 two real numbers. Find a harmonic function u on the strip {y 0 < Im z < y 1 } with u = 0 on {Im z = y 0 } and u = 1 on {Im z = y 1 }. 2. Let 0 θ 0 < θ 1 < 2π. Find a harmonic function u on the wedge {z = ρe iθ, θ 0 < θ < θ 1 }, satisfying u = 0 on {arg z = θ 0 } and u = 1 on {arg z = θ 1 }. 3. Find a harmonic function on the unit disk satisfying the following boundary conditions: (a) u D (e iθ ) = sin 2 θ. (b) u D (e iθ ) = 3 + sin(2θ). (c) u D (e iθ ) = cos 3 θ. Material used for this lecture: [Taylor],

37 37 Lecture 12/13 - More Laplace problems using conformal equivalences and particular solutions Example 20. Using the solution v(x, y) = π tan 1 y x v = 0 ({Rz > 0}), v Arg z= π 2 to the problem = 0, v Arg z= π 2 = 2, Find a function u harmonic on the right-half plane, satisfying the boundary conditions u(0, y) = 2 if y 1, u(0, y) = 0 if y > 1. Answer: Let h a LFT mapping the triple ( i, 0, i) to (0, i, ), that is, you may find h(z) = i z+i z i. It maps the imaginary axis to itself and h(1) = 1 so the right-half plane is mapped onto itself. The function v 1 = u h 1 is harmonic on the RHP and solves the boundary condition v 1 = 0 on { it, t (0, + )} and v 1 = 2 on {it, t (0, + )}. By uniqueness principle, it must equal to v above, i.e. v = u h 1, that is to say, u = v h. So the answer is u(x, y) = π tan 1 Im h(x, y) Rh(x, y) = π tan 1 ( 1 x 2 y 2 Remark 5. The choice we have in the LFT we pick does not change the uniqueness of the solution to our problem. Indeed, the choice of the image of the point 0 by the LFT we picked does not matter as long as it is of the form iy with y > 0: In that case, the new LFT would be a multiple of h(z) via a real scalar, so that the ratio Im h(x,y) Rh(x,y) 2x ). in the final result would not be changed. Example 21. Call v(x, y) the solution of Example 19. Let 0 < t < 2π. Find a Linear Fractional Transformation h such that the function u = v h is harmonic on D and satisfies the boundary conditions u = 2 on {e iθ, θ (0, t)} and u = 0 on {e iθ, θ (t, 2π)}. Answer: Similar to the previous example, let h a LFT mapping the unit disc to itself. We prescribe that h maps the triple (1, e i t 2, e it ) to (1, i, 1). Once h is computed, with an argument similar to the previous example, the final solution is u = v h, where v is the solution of example 19. Example 22. Using the solution of Example 21 and the superposition (linearity) principle, find a harmonic function u on D satisfying the boundary conditions: u = 1 on {e iθ, θ (0, π 2 )}, u = 2 on {e iθ, θ ( π 2, π)} and u = 0 on the lower-half unit circle. Answer: Let us define C 1 = {e iθ, 0 < θ < π 2 } and C 2 = {e iθ, π 2 < θ < π}. Using the superposition principle, we write u = u 1 + u 2, where u 1 solves: and u 2 solves u 1 = 0 (D), u 1 C1 = 1, u 1 D C1 = 0, u 2 = 0 (D), u 2 C2 = 2, u 2 D C2 = 0, In fact, in this particular case, the second solution u 2 can be transported to the first one via a 90-degree rotation and a scaling by 2, so u 2 (z) = 2u 1 (iz) (i.e. u 2 (x, y) = 2u 1 ( y, x)).! The first solution u 1 is obtained via the method in example 21.

38 38 Example 23. Let 0 < t 1 < < t n < 2π and a 1,..., a n real numbers. Think of a way of generalizing the approach above to find a harmonic function u on D satisfying the boundary conditions: for 1 k n, u = a k on the arc-circle {e iθ, θ (t k, t k+1 )} (where we have defined for convenience, t n+1 = t 1 + 2π). Answer: write u = n k=1 a ku k, where, if we denote C k = {e iθ, t k < θ < t k+1 }, u k solves the problem u k = 0 (D), u k Ck = 1, u k D Ck = 0, These are all computible from example 21 and, using the superposition principle, the function u = n k=1 a ku k solves the initial problem. Definition 13. A lunar domain is a domain whose boundary consists of portions of two given circles in C. Lunar domains owe their name to some crescent-looking cases where both circles in C are actual circles: for instance pick the circles { z = 1} and { z 1 = 1} and consider the portion of the plane containing the point z = 1 2. In this case, the other three portions of the plane whose boundary is delimited by both circles are also lunar domains. Other lunar domains can be bounded by an arc circle and a flat boundary, or two flat boundaries (in which case, they may give sectors or strips). It turns out that these lunar domains are all conformally equivalent to sectors, which are domains where we can easily find harmonic solutions with constant boundary values on each ray based on the argument function. Example 24. Let Ω the lunar domain consisting of the upper-half-plane minus the closed unit disk: Ω = {Im z > 0} D. Its boundary consists of a flat part Γ 1 and a round part Γ 2. Find a function u harmonic on Ω such that u Γ1 = 0 and u Γ2 = 2. Answer: The LFT h(z) = z 1 z+1 sends 1 to 0 and 1 to. The two portions of circle forming the boundary must be sent, via h, into circles of C passing through 0 and, i.e. they must be radial rays. Computing h(2) = 1 3, we conclude that h(γ 1) is a radial line of direction θ = 0. Computing h(i) = i, we conclude that h(γ 2 ) is the radial line of direction θ = π 2. The transported problem on the image domain is: find a harmonic function on the top-right quadrant such that v(x, 0) = 0 for x > 0 and v(0, y) = 1 for y > 0. The solution to this is v(x, y) = 2 π Arg (z) = 2 π tan 1 y x. The final solution is then u = v h, that is to say, u(x, y) = 2 Im h(x, y) tan 1 = 2 ( ) 2y π Rh(x, y) π tan 1 x 2 + y 2. 1

39 39 Exercises for lecture 12-13: 1. Let C 1 = {e iθ, θ ( π 4, π 4 )} and C 2 = {e iθ, θ ( 3π 4, 5π 4 )}. Find the function u, harmonic inside the unit disc D, satisfying the boundary condition u C1 = 1, u C2 = 1, u D C1 C 2 = Let the lunar domain U = D 1 (1 + i) {y > x + 1}. (a) Draw U, identify the two tips of the crescent, and propose a LFT mapping U to a sector. (b) Describe completely the domain h(u) and draw it. (c) Using (b), find a function u harmonic on U such that u = 1 on the straight edge of U and u = 1 4 on the circular edge of U.

40 40 Lecture Homotopy and simple connectedness Definition 14. Let u C an open, connected set. Two closed curves γ 0 : [0, 1] U and γ 1 : [0, 1] U are called homotopic in U if there exists a continuous function h : [0, 1] 2 U satisfying h is called a homotopy from γ 0 to γ 1. h(0, t) = γ 0 (t), h(1, t) = γ 1 (t), t [0, 1], h(s, 0) = h(s, 1), s [0, 1]. The homotopy h can be seen as a continuous deformation from γ 0 to γ 1. For s [0, 1] fixed, one may regard t h(s, t) as a closed curve, which transitions continuously from γ 0 to γ 1 as s goes from 0 to 1. Figure 4: Example of two homotopic paths in C. γ 0 (t) = e 2πit and γ 1 (t) = 1+i+(2+cos(2πt))e 2iπt for t [0, 1]. The homotopy here is given by h(s, t) = (1 s)γ 0 (t) + sγ 1 (t). Remark 6. Being homotopic in U is an equivalence relation among closed paths in U. Two paths are homotopic if one can be continously deformed into the other. Example Let γ 0 (t) = e 2iπt and γ 1 (t) = e 4iπt for t [0, 1]. Then γ 0 and γ 1 are homotopic in C but not in C {0}, we will prove this after the Proposition below. 2. If U is convex, then any two closed curves γ 0,1 : [0, 1] U are homotopic in U, via the homotopy h(s, t) = (1 s)γ 0 (t) + sγ 1 (t). h is obviously continuous and valued in U because h(s, t) [γ 0 (t), γ 1 (t)] U. Points are examples of (constant) closed curves given by γ : [0, 1] C such that γ(t) = z 0.

41 41 Proposition 16. If γ 0 and γ 1 are homotopic in U and f is analytic on U, then γ 0 f(z) dz = γ 1 f(z) dz. Proof. Let h a homotopy from γ 0 to γ 1, suppose here that h has continuous second partial derivatives here. Call I(s) = 1 0 f(h(s, t)) th(s, t) dt. And let us show that di/ds = 0, so that I(0) = I(1) which will conclude the proof. We compute di ds = 1 0 s (f(h(s, t)) th(s, t)) dt = 1 Using the fact that s t h = t s h, we can rewrite this as di ds = = hence the result. 0 [f (h(s, t)) s h(s, t) t h(s, t) + f(h(s, t)) s t h(s, t)] dt. [f (h(s, t)) s h(s, t) t h(s, t) + f(h(s, t)) t s h(s, t)] dt t (f(h(s, t)) sh(s, t)) dt = f(h(s, 1)) s h(s, 1) f(h(s, 0)) s h(s, 0) = 0, We can now show Example 1: γ 0 and γ 1 are homotopic in C because C is convex. By contradition, if they were homotopic in C {0}, then considering the function f(z) = 1 z, analytic on C {0}, this would mean that dz γ 0 z = dz γ 1 z, but direct computation shows that the LHS equals 2πi while the RHS equals 4πi. Hence γ 0 and γ 1 are not homotopic in C {0}. Definition 15. A set U C is called simply connected if every closed curve in U is homotopic to a point. By transitivity of being homotopic, when U is simply connected, any two closed curves in U are homotopic to one another in U. Remark 7. When γ is a constant, since γ (t) = 0 for any f, γ f = 0. Therefore, when f is any analytic function on U simply connected, for any closed curve γ, we have γ f = 0. Example 26. Any convex (or, weaker yet, star-shaped ) set is simply connected, in particular C and open discs. C {0} is not simply connected by virtue of Example 25. Properties of analytic functions on simply connected sets. Proposition 17. If U is simply connected, then every analytic function admits an antiderivative. Proof. Let f : U C analytic on U simply connected. Fix z 0 in U and define g(z) = γ z0 f(w) dw,,z where γ z0,z is any curve with endpoints z 0, z. By virtue of the previous proposition, this definition does not depend on the choice of curve: indeed, if γ 1 and γ 2 are two curves with endpoints z 0, z, then γ 1 γ 2 is a closed curve in U, therefore γ 1 γ 2 f(w) dw = 0, that is, γ 1 f(w) dw = γ 2 f(w) dw. It is then easy to show that g (z) exists and equals f(z).

42 42 Proposition 18. If U is simply connected, every harmonic function admits a harmonic conjugate. Proof. Let u harmonic in U. Define f = u x iu y. Then the real and imaginary parts of f satisfy the Cauchy-Riemann equations on U, therefore f is analytic on U. By the previous proposition, f admits and antiderivative g such that g = f. In particular, if we set v = Im(g), we have that g = v y + iv x, so that the equality g = f entails u x = v y and v x = u y on U, which exactly means that v is a harmonic conjugate for u. lecture 15 (2/11) Definition 16. An analytic function f : U C is said to have: an analytic logarithm on U if there exists h : U C analytic such that f(z) = exp(h(z)). an analytic squareroot on U if there exists g : U C analytic such that f(z) = g 2 (z). Remark If U = D and f(z) = z, then f admits no analytic logarithm on D. This is because any definition Log z needs to have a branch cut in the form of a curve joining 0 and. 2. If U = D 1 (2), then Log z = ln z + i arg ( π,π) (z) is an analytic logarithm of f(z) = z on U. Theorem 19. If U is simply connected, then any analytic, non-vanishing function on U admits (i) an analytic logarithm, and (ii) an analytic squareroot. Proof. (i) If f is analytic, non-vanishing on U, then the function u(z) = ln f(z) is harmonic on U. By the previous proposition, u admits a harmonic conjugate v on U. Denoting h = u + iv, h is clearly analytic, therefore so is the function fe h. Moreover, this function is such that for every z U, f(z)e h(z) = f(z) e Re(h(z)) = f(z) e u(z) = f(z) f(z) 1 = 1, so the modulus of fe h achieves a local maximum inside of U, which by virtue of the Maximum Modulus principle 4 implies that fe h is a constant, call it c = e b for some b C (think of why c 0). Then the function z h(z) + b is analytic on U and such that f = e h+b, that is, h + b is an analytic logarithm for f on U. (ii) Now that (i) is true, let h(z) an analytic logarithm for f, then g(z) = exp h(z) 2 is the composition of analytic functions and satisfies g 2 (z) = f(z), so g is an analytic squareroot for f. 4 Maximum modulus theorem: If f is analytic on a connected open set U and f has a local maximum at z 0 U, then f is constant on U. See [Taylor, Theorem 3.5.2]

43 43 Exercises for lectures 14-15: 1. Prove that being homotopic in U is an equivalence relation among closed curves in U. 2. Find a homotopy h(s, t) from the boundary of the unit square {z = e it max( cos t, sin t ), t [0, 2π)} = [ 1, 1] 2, to the unit circle {z = e it, t [0, 2π)} = { z = 1}. 3. An open set U is star-shaped with respect to z 0 U if and only if for every z U, the segment [z 0, z] = {(1 t)z 0 + tz, t [0, 1]} is included in U. Prove that a star-shaped domain is simply connected. 4. Are the following sets simply connected? (a) The open right-half plane with the segment [1, + ) removed. (b) The open right-half plane with the segment [1, 2] removed. (c) The open right-half plane with both segments [1, + ) and (i, 2 + i] removed. 5. If U can be written as U = V W with both V and W simply connected and V W, is it true that U is simply connected? If not, suggest without proof an additional requirement on V W in order to make U simply connected. 6. Suppose U and V are two open domains which are homeomorphic, i.e. there exists a bijection h : U V such that both h : U V and h 1 : V U are continuous. Prove that U is simply connected if and only if V is simply connected. 7. Check by direct computation that u(z) = ln f(z) is harmonic wherever f is analytic nonvanishing. 8. For n N, an analytic function f : U C admits an analytic n-th root if there exists an analytic function h : U C such that f(z) = (h(z)) n for every z U. Prove that if f admits an analytic logarithm on U, then for every n N, f admits an analytic n-th root on U. Remark 9. The conclusion of ex. 5 also holds if U and V are conformally equivalent, as being conformally equivalent is a stronger statement than being homeomorphic.

44 44 Lecture 16 - Compactness and Montel s theorem Let us recall the following facts on compact sets of C. K C is compact if and only if any of the properties below is satisfied: for every open covering O = i O i of K, there exists a finite subcovering of O covering K. every sequence {z n } n in K has a subsequence that is convergent in K. K is closed and bounded. The one-dimensional analogue is the Bolzano-Weierstrass theorem, stating that a sequence of real numbers inside a finite interval [a, b] has a subsequence converging to a limit in [a, b], in other words, closed, finite intervals are compact. As we see, compact sets are useful for generating convergent sequences out of any sequence, provided that this sequence is in a compact set. Compactness is a powerful tool in proving existence theorems provided that we have a good understanding of what sets are compact given a certain notion of convergence. Below we explain how to extend this concept to families of analytic functions (instead of points in C), and how to characterize families of analytic functions (Montel s theorem) with normally convergent subsequences. The theorem below says that it takes very little for a sequence of analytic functions to have a normally convergent subsequence: the functions only have to be uniformly bounded on U by the same constant. Theorem 20 (Montel s theorem). Let f n : U C a sequence of analytic functions on U open. Suppose there exists M 0 such that f n (z) M for every z U and n N, then f n has a subsequence converging normally on U. Example 27. For any open set U C, a sequence {f n } n satisfying f n : U D 1 (0) for every n N is uniformly bounded by 1, so it has a normally convergent subsequence. We now focus on the proof of Montel s theorem, which requires a few steps. Proof of Montel s theorem. Let f n : U C a sequence uniformly bounded by M on U. Step 1: construct a candidate for a normally convergent sequence. The set of points in U with rational coordinates, i.e. U (Q + iq), is countable and dense in U, so let us consider an enumeration z 1,..., z n,... of points in U (Q + iq). The sequence (of complex numbers) {f n (z 1 )} n is bounded by M so there exists a converging subsequence of it, call it {f 1n (z 1 )} n. Now the sequence {f 1n (z 2 )} is bounded as well, and thus has a convergent subsequence, call it f 2n (z 2 ) (note that the sequence f 2n (z 1 ) is a subsequence of the convergent sequence f 1n (z 1 ) so it converges as well). Proceeding inductively, we can construct for every k a subsequence f kn

45 45 of f n such that the sequences {f kn (z j )} n converge for each j k. Defining g n := f nn, g n is a subsequence of f n such that {g n (z j )} n converges for every j. Step 2: a uniform estimate. Let w U and r > 0 such that D 2r (w) U, then we establish that f n (z) f n (z ) M r z z, z, z D r (w), n. Indeed, for any z D r (w), the disk D r (z) is included in the domain of analyticity of f n, so Cauchy s estimate 5 implies that f n(z) M r. Then the identity is derived using an ML estimate: f n (z) f n (z ) = f n l([z, z ]) max f n z z M [z,z ] r. (6) [z,z ] Step 3: g n converges uniformly on closed disks. Let w U and r > 0 as in step 2. We will show that {g n } n is uniformly Cauchy on D r (w), that is ε > 0, N > 0, n, m N, z D r (w), g n (z) g m (z) < ε. Let ε > 0. Define δ = rε 3M > 0, then there exists finitely many z 1,..., z p with rational coordinates such that D r (w) p j=1 D δ(z j ) (this is because, since D r (w) is compact and {z n } is dense in it, it is covered by {D ρ (z j )} j N, out of which we can extract a finite subcover). Since the sequences {g n (z j )} n all converge for j = 1... p, we can find an N such that n, m N, j = 1... p, g n (z j ) g m (z j ) < ε 3. (7) (this is only possible because the z j s chosen are finitely many!). Then for z D r (w), let z j with rational coordinates such that z z j < δ. Then for n, m N, we bound g n (z) g m (z) = g n (z) g n (z j ) + g n (z j ) g m (z j ) + g m (z j ) g m (z) M r δ + ε 3 + M r δ = ε 3 + ε 3 + ε 3 = ε, where the first and third terms are bounded using (6) and the middle term is using (7). Step 4: uniform convergence on any compact set. This part is left as an exercise. Let K a compact subset of U, then show that K is covered by finitely many discs over each of which {g n } n converges uniformly. 5 Cauchy s estimate: If f : U C is analytic on U open and z 0 U, ρ > 0 are such that D ρ(z 0) U, then we have f (z 0) 1 ρ max z z 0 =ρ f(z).

46 46 Exercises for lecture 16: Using Montel s theorem, for each sequence below, prove that {f n } has a subsequence which converges normally in D. 1. f n (z) = cos(nz). [Warning: cos z is not bounded by 1 for every complex z!] 2. f n (z) = e iπn 2 z. 3. f n (z) = 3z+sin(2n) 5+2z cos n. 4. Let w n a sequence in D and define f n (z) = n z w k k=1 1 w k z. [First show that for any z D, z w k 1 w k z D. One way to do this is to show that z z w k 1 w k z is an LFT mapping the unit circle to itself and its interior to itself]. Material used for this lecture: [Taylor, Theorem 6.4.2] (some details in the proof are missing)

47 47 Lecture 17 - Conformal Automorphisms of the unit disk We now want to characterize the conformal automorphisms of the unit disk. Let us define Aut D = {f : D D, f analytic, bijective}. A straighfoward observation is that Aut D is a group under composition: the identity map is in it and composition and inverses of bijective, analytic functions are bijective, analytic. Our goal is to characterize Aut D in a simple way. Namely we want to arrive at the following result: for w D, let us define the Möbius transformation h w (z) = z w 1 wz, and for θ [0, 2π), let us define R θ (z) = e iθ z. We will show that Aut D is made exactly by functions of the form h(z) = eiθ z w 1 wz = R θ h w (z) for some θ [0, 2π) and some w D. Before proving this, some preliminaries. Lemma 21 (Schwarz s lemma 6 ). Let f : D D analytic such that f(0) = 0. Then f(z) z for all z D and f (0) 1. Moreover, if f (0) = 1, then f(z) = cz for some constant c such that c = 1. Proof. Since f(0) = 0 and f is analytic on D, f can be written as f(z) = zg(z) with g analytic on D. Let us show that g(z) 1 for every z D. For any 0 < R < 1 and z such that z = R, we have g(z) = f(z) z = f(z) R 1 R. Then by the maximum principle, we have g(z) max D R (0) 1 R, z D R(0). Sending R to 1, we are able to show that g(z) 1 for every z D. Therefore we have f(z) = z g(z) z for every z D. and f f(z) f(0) (0) = lim z 0 z 0 = lim z 0 g(z) = g(0), so f (0) = g(0) 1 as well. Now suppose that f (0) = 1. Then g(0) = 1, and since g(z) 1 throughout D, g has a local max at z = 0, hence g must be constant. Certainly, this constant c must satisfy c = 1 and for every z D, f(z) = zg(z) = cz. Theorem 22 (Theorem in [Taylor]). Let f a conformal automorphism of D with f(0) = 0, then f(z) = cz for some constant c such that c = 1. Proof. Based on the previous result, all we need to show it that f (0) = 1 and the conclusion will follow. Since f is bijective, then f 1 exists and we have for every z D, f 1 (f(z)) = z. Differentiating this relation, we obtain f (z)(f 1 ) (f(z)) = 1, in particular at z = 0, we get f (0) (f 1 ) (0) = 1. But both functions f and f 1 satisfy the hypotheses of Schwarz s lemma, so we have f (0) 1 and (f 1 ) (0) 1. This can only be achieved if f (0) = (f 1 ) (0) = 1. Hence the result. 6 see e.g. Lemma in [Taylor]

48 48 Theorem 23 (Characterization 7 of Aut D). h is a conformal automorphism of the unit disc (i.e. h : D D is bijective and conformal) if and only if it is of the form h(z) = e iθ h w (z) for some numbers w D and θ [0, 2π). Proof. ( = ) The first thing to notice is that such candidates are indeed in Aut D. h(z) = e iθ h w (z) is the composition of a rotation about 0 with the map h w (z). It is easy to check that the rotation is an element of Aut D. Now on to h w, we know it is a Möbius transformation, of which we prove the following facts: h w maps the unit circle to itself: indeed, if z = 1, then h w (z) = z w 1 wz = 1 z w z z w = z w z w = 1, since z 1 = z whenever z = 1. h w maps the interior of the unit circle to itself: indeed, h w (0) = w D. h w is conformal since it is a Möbius transformation. Therefore, for every w D and θ [0, 2π), z e iθ h w (z) is a conformal automorphism of D. ( = ) Suppose h Aut D and let us split cases. Either h(0) = 0, in which case Theorem 22 concludes that h(z) = e iθ z for some θ [0, 2π). Or h(0) 0, in which case w = h 1 (0) 0. Consider the function g = h h w = h h 1 w. We have g Aut D as the composition of two elements of Aut D, moreover g(0) = h(h 1 w (0)) = h(w) = 0, so by the previous case, it must be that g(z) = e iθ z for some θ [0, 2π). Then h h 1 w = R θ, that is to say, h = R θ h w. This ends the proof. Remark 10. So Aut D = {R θ h w, w D, θ [0, 2π)}. This makes Aut D a three real-parameter group. It s a subgroup of the Möbius transformations (a six real-parameter group after we normalize them) which map the unit circle to itself. Corollary 24. Suppose U C is an open set that is conformally equivalent to D, that is to say, there exists h : U D conformal and bijective. Then the conformal automorphisms of U are described by Aut U = {h 1 g h, g Aut D} In particular, we can transport the automorphisms of the unit disk to any domain that is conformally equivalent to D. Examples exploiting this fact for artistic purposes can be found in [Bulatov]. 7 see e.g. Theorem in [Taylor]

49 49 Exercises for lecture 17: 1. Prove Corollary Fixing w D, prove that h 1 w = h w. 3. Prove that the characterization of Theorem 23 is unique: if for every z D, e iθ h w (z) = e iθ h w (z) for θ, θ [0, 2π) and w, w D, then θ = θ and w = w. [Hint: first look at images of well-chosen points] 4. Prove that for any w D and θ [0, 2π), h w R θ = R θ h e iθ w. 5. Recall the classification of Möbius transformations from Lecture 9. Check the following facts and, for all cases, roughly sketch a typical phase portrait of the situation. (a) All transformations of the form R θ are either elliptic or parabolic. (b) All transformations of the form h w are hyperbolic if w 0, compute their fixed points. (c) If g Aut D has a fixed point w D {0}, i.e. g(w) = w, show that there exists θ [0, 2π) such that g = h w R θ h w. In particular, show that g is either elliptic or parabolic, using (a).

50 50 Lecture 18 - The Riemann Mapping Theorem We now have all the tools required to prove the Riemann mapping theorem, stating that if U C is simply connected, then there exists a conformal equivalence between U and the unit disc D = D 1 (0). The proof requires a few steps, which we now break down. Let U C an open, simply connected domain, and let z 0 U fixed. Denote by F the family of functions We will show the following: Lemma 25. The set F is non-empty. F = {f : U D analytic, one-to-one, f(z 0 ) = 0}. Proof. Since U C, let λ U fixed. Then the function f(z) = z λ is analytic, one-to-one and non-vanishing on U, therefore it admits an analytic squareroot g on U. g is clearly one-to-one (if it wasn t, then neither would be f). We claim g(u) is open. Indeed, since g is one-to-one, then it is bijective onto g(u), and the relation g(z) 2 = z λ implies that the inverse of g admits the expression g 1 (w) = w 2 + λ for w g(u), which is clearly analytic, hence continuous. Therefore, since U is open and g 1 is continuous, g(u) = (g 1 ) 1 (U) is open. Since g(u) is open, there exists D r (w 0 ) g(u) for some r > 0, but then since g 2 = f and f is one-to-one, we necessarily have D r ( w 0 ) g(u) =, otherwise if two opposite points are in the range of g, then their images by f are the same, contradicting injectivity. This implies the relation g(z) + w 0 > r, z U, r therefore the function p(z) = g(z)+z 0 is analytic, one-to-one on U such that p(z) < 1, so p : U D. If p(z 0 ) = 0, then p F, and if p(z 0 ) 0, then h p(z0 ) p F, where h p(z0 ) = z p(z 0) is a conformal 1 p(z 0 )z automorphism of the unit disc taking p(z 0 ) to 0. In the next lemma, we will use automorphisms of the unit disc h w (z) = z w 1 wz, in particular the following Lemma 26. h w(0) = 1 w 2 and h w(w) = 1 1 w 2. Proof. Straightforward computation. Lemma 27. If f F and f is not onto, then there exists g F satisfying g (z 0 ) > f (z 0 ).

51 51 Proof. Let f F not onto and let w / f(u), then h w f is a non-vanishing analytic function on U, therefore it admits an analytic squareroot q such that q 2 = h w f. q does not vanish either so q(z 0 ) = λ 0 and λ 2 = (q(z 0 )) 2 = h w (f(z 0 )) = h w (0) = w. Let us set g = h λ q and show that g fulfills the lemma. We compute g (z 0 ) = h λ (q(z 0))q (z 0 ) = h λ (λ)q (z 0 ) = Differentiating the relation q 2 = h w f at z 0, we arrive at 1 1 λ 2 q (z 0 ). 2q(z 0 )q (z 0 ) = h w(f(z 0 ))f (z 0 ) = h w(0)f (z 0 ) = (1 w 2 )f (z 0 ) = (1 λ 4 )f (z 0 ). Concluding, g (z 0 ) = 1 1 λ 4 1 λ 2 f (z 0 ) = 1 + λ 2 f (z 0 ). 2λ 2λ The proof is complete by checking that whenever λ 1, 1+ λ 2 2 λ > 1. Theorem 28 (Riemann mapping theorem). There exists an element h F that is onto. particular, h is a conformal equivalence between U and D. In Proof. The family F is uniformly bounded, so by Montel s theorem, it is normal. Let us denote m = sup f (z 0 ). f F Since conformal mappings have non-vanishing derivatives so m > 0, and m is finite by using Cauchy estimate (there exists r > 0 such that D r (z 0 ) U, then for every f F, f (z 0 ) max z z0 =r f(z) R 1 R ). From the characterization of the supremum, there exists a sequence f n F such that lim n f n(z 0 ) = m. Using Montel s theorem, since the sequence f n is uniformly bounded by 1 on U, f n admits a subsequence {f nk } k which converges normally to a function h : U C. We claim that h F. Clearly, h(z 0 ) = lim k f nk (z 0 ) = lim k 0 = 0. h is analytic as a uniform limit of analytic functions. h (z 0 ) = lim k f n k (z 0 ) = m. Since m 0, h is not a constant. Then by Hurwitz s theorem, since all f nk s are one-to-one, then h must be one-to-one. We claim that h is onto. By contrapositive of the previous lemma, we have h (z 0 ) g (z 0 ) for every g F, so h is onto.

52 52 Exercises for lecture 18: 1. Let U C open, simply connected such that its complement contains a disk D r (w) for some r > 0 and w C. Then suggest a simpler proof that F using linear fractional transformations. 2. Prove that h w(0) = 1 w 2 and h w(w) = 1 1 w 2 when h w (z) = z w 1 wz. 3. Although C is simply connected, prove that it is not conformally equivalent to D (use Liouville s theorem). 4. Is an open rectangle conformally equivalent to D? 5. Prove that if f is analytic and satisfies f(z) < 1 on the right-half plane, and if f(1) = 0, then f(z) z 1 z+1 for all z in the right-half plane. [Hint: Schwarz s lemma and a conformal equivalence from right-half plane to D]

53 53 Lecture 19 - Introduction to complex dynamics We first start with real dynamics, that is, let F : R R a continuously differentiable function, x 0 R and consider the sequence with initial term x 0 and induction relation x n+1 = F (x n ), so that x n = F n (x 0 ), where we denote F n (x 0 ) the n-fold compound F F F (Caution: F n (x 0 ) is NOT (F (x 0 )) n!). n plays the role of a (discrete) time variable. Definition 17. x 0 is called the seed and the sequence {F n (x 0 )} n is called the orbit of x 0. Example 28. Consider the sequence starting at x 0 > 0 and x n+1 = αx n + β, where α > 1 and β R. By computing the first few terms, we can show by induction that the general term takes the form x n = α n x 0 + αn 1 α 1 β, so that we have an explicit expression which allows us to infer the longterm behavior of the sequence. It turns out that for most subsequent cases, such explicit formulas are no longer possible, hence the need for more qualitative methods to study dynamical systems. Example 29. Another example is the logistic equation x n+1 = kx n (1 x n ), where x n = x(n) denotes a population at time n, a simple model expressing the fact that the variation of a population over one unit of time is proportional to itself, while that variation can become negative (due to the term 1 x n ) when the population passes a certain threshold, modelling the need to compete for resources if the population becomes overwhelming. Depending on the value of k, this model can display very different dynamical features, of comparable complexity as what we will see when studying the quadratic family F c (x) = x 2 + c in the next lectures. Example 30. Iterating a function appears naturally in Newton s method, whose purpose is to find solutions of an equation f(x) = 0 for f given, by setting up the iterative scheme x n+1 = x n f(xn) f (x n). Since we have no explicit method a priori to find the zeros of a general function (except for polynomials of order up to 5), the recursion relation above means that, starting at a seed x 0, we pick x 1 as the unique zero of the linear function approximating f at x 0 (i.e. the Taylor expansion p(x) = f(x 0 ) + (x x 0 )f (x 0 )), then we repeat this procedure at x 1 and so on... In this case, the map that is being iterated over is F (x) := x f(x) f (x), called the Newton map of f. The study of Newton s algorithm was one of the main motivations for studying iteration of functions [A]. Typical questions that one may ask are: 1. What is the long-term behavior of the orbits of a given dynamical system x n+1 = F (x n )? Possible answers are: convergent, divergent, periodic, asymptotically periodic, none of the above (leading to rather erratic trajectories) 2. Given a one-parameter family of dynamical systems x n+1 = F c (x n ) (or the k-dependent family from Example 2), how does the answer to question 1 vary in terms of the values of c? This is the question behind bifurcation theory. In particular, we will spend some time studying the family F c (x) = x 2 + c for c R. 3. These questions can then be generalized to the complex plane, where F : C C and the dynamic sequence z n+1 = F (z n ) is now complex-valued.

54 54 Fixed points. We start with the following observation. Here and below, we fix F a continuously differentiable function F. Theorem 29. If an orbit x n = F n (x 0 ) converges to x R, then x is a fixed point of F, i.e. F (x ) = x. Proof. F (x ) = F ( lim n x n) = lim n F (x n) = lim n x n+1 = x, where we have used the continuity of F at x. Hence we see that a dynamical system has convergent orbits only if it has fixed points, this also gives us an obvious (constant) orbit {x n = x } n for this dynamical system. The next notion then concerns stability of these constant orbits: what happens if we perturb the seed x 0 a little bit away from a fixed point x, does the orbit still converge to x? Definition 18. A fixed point of F is called attracting if F (x ) < 1. super attracting if F (x ) = 0. repelling if F (x ) > 1. neutral if F (x ) = 1. The next theorem justifies the terminology attracting above. Theorem 30. If x is attracting, then there exists ε > 0 such that for every x 0 (x ε, x + ε), the orbit of x 0 converges to x. Proof. By continuity of F, since F (x ) < 1, there exists ε > 0 and 0 < β < 1 such that F (x) < β for every x such that x x < ε. Then for every x, y (x ε, x + ε), we have F (x) F (y) β x y. In particular, for a seed x 0 (x ε, x + ε), we have x 1 x = F (x 0 ) F (x ) < β x 0 x. By induction, one can see that x n x < β n x 0 x and since β < 1, then by Squeeze prnciple, lim n x n x = 0. Remark 11. Similarly to this proof, we can understand the concept of repelling, in which case one would have instead F (x) F (y) > β x y near x for some β > 1, thereby preventing any sequence starting near x to converge to x. Remark 12. If x is superattracting, then F (x) F (x ) β(x x ) 2 for x near x, so that convergence to x becomes quadratic instead of linear, i.e. much faster. The conclusion is not as straightforward when a fixed point is neutral, as we will see below.

55 55 Graphical Analysis. We now introduce a brief method to easily draw trajectories. The idea is to plot the function F together with the line {y = x}. Drawing the orbit of a given x 0 can be done by following the strategy below: start from x 0 on the real axis, move vertically toward the graph of f, then move horizontally toward the line {y = x} and repeat the process. The sequence x n can be read as all the successive abscissae of the points hit on the line {y = x}. See typical situations on Fig. 5. Figure 5: Examples of graphical analyses in the case of an attracting point (left) with F (x) =.05x x + 1 and a repelling point (right) for F (x) = 2x.03x 2. Remark on neutral fixed points: Using this tool, it is easy to realize that neutral fixed points can exhibit all possible behaviors: pick a function F whose graph is tangent to {y = x} at some x. To the left of x, if the graph of F lies above the curve {y = x}, graphical analysis will show that nearby seeds will converge to x. If the graph of F lies below the curve {y = x}, nearby seeds will diverge away from x. To the right of x, the opposite situation happens. For each situation above, since there are functions whose graphs can achieve either situation, the overall behavior of a neutral point can be: attracting, repelling, or attracting on one side and repelling on the other.

56 56 Exercises for lecture 19: 1. Depending on the values of α, β in Example 28, discuss the long-term behavior of the sequence x n. Consider α R and β R. 2. Show that when f is a continuously differentiable function and F its Newton map, (a) all zeros of f are fixed points of F. (b) the fixed points of F where f does not vanish are superattracting. 3. (a) Show that the iteration x n+1 = 1 2 (x n + a x n ) is Newton s method for the polynomial f(x) = x 2 a. (b) Discuss heuristically (with graphical analysis) the convergence of the sequence {F n (x 0 )} n in the case a > 0 (does it converge? what is its limit?) depending on the value x 0 of the initial seed. (c) Draw a graph of F (x) = 1 2 (x+ 2 x ) and the line {y = x}, and draw a few typical trajectories for this system. x n+1 (extra) Prove that for any seed, the orbit sequence satisfies a x n prove rigorously that lim n x 2 n a = 0. Conclude based on the sign of x 0. x n a x n and 4. Set up Newton s algorithm for the polynomial f(x) = x 2 +1: roughly sketch the Newton map of f and draw a few trajectories of the corresponding dynamical system. What seems to be happening? Material used for this lecture: [DK] Dynamics of simple maps, R. Devaney.

57 57 Lecture 20 - Continuation of one-dimensional dynamics. In the previous lecture, we emphasized the importance of fixed points as potential finite limits of orbits, and defined concepts related to stability of orbits: if x is a fixed point of F, this fixed point is classified as superattracting, attracting, neutral or repelling if, respectively, F (x ) = 0, 0 < F (x ) < 1, F (x ) = 1 or F (x ) > 1. Definition 19. For x a fixed point of F : R R, the basin of attraction of x, denoted A F (x ) is the set of all seeds whose orbits converge to x, that is, A F (x ) = {x R : lim n F n (x) = x }. Remark 13. Using this terminology, we have proved in previous lecture that if x is attracting or superattracting, then there exists ε > 0 such that (x ε, x + ε) A F (x ). Example 31. Let F (x) = x 2. Find all fixed points of F and classify them. By graphical analysis, deduce the long-term behavior of all orbits. Solution: F (x) = x is equivalent to x(x 1) = 0 so the fixed points are 0 and 1. F (0) = 0 so x = 0 is superattracting and F (1) = 2 so x = 1 is repelling. Graphical analysis arrives at the following conclusions: if x ( 1, 1), the orbit of x converges to 0. if x {±1}, the orbit of x converges to 1. if x > 1, then lim n F n (x) = +. In this case, we have A F (0) = ( 1, 1), A F (1) = {±1}, which tells us that basins of attraction can be topologically quite diverse (A F (0) is open and not closed, A f (1) is closed and not open). Moreover, if we extended F as a meromorphic function on the extended complex plane, it would make sense to say that is another fixed point of F (F ( ) = ) so we could talk of the basin of attraction of and say that A F ( ) = (, 1) (1, + ). Example 32. Same instructions as above for the function F (x) = x 2 1. Solution: Searching the fixed points of F yields x = x ± := 1± 5 2. Then F (x + ) = 2x + = > 1 and F (x ) = 1 5 < 1 so both fixed points are repelling. By graphical analysis, we see that if x > x +, we again have lim n F n (x) = +, if x = ±x +, then lim n F n (x) = x +, and if x = ±x, lim n F n (x) = x. On the other hand all other points seems to generate an orbit which eventually alternates between the values 0 and 1. This motivates the definition of periodic orbits and their stability. Let us define these concepts and return to out example. Definition 20. A point x is k-periodic is F k (x) = x and F j (x) x for every 0 < j < k.

58 58 When a point x 0 is k-periodic, calling x j = F j (x 0 ) for 0 < j < k, its orbit looks like We call such an orbit a k-cycle. x 0, x 1,..., x k 1, x 0, x 1,..., x k 1, x 0, etc... Theorem 31. For a k-cycle (x 0, x 1,..., x k 1 ), we have the property that each point x j is k-periodic, moreover, (F k ) (x 0 ) = (F k ) (x 1 ) = = (F k ) (x k 1 ). Calling λ this number, it is called the eigenvalue or multiplier of the cycle. Proof. Direct application of the chain rule. Definition 21. A cycle is, respectively, super-attracting, attracting, neutral or repelling if its eigenvalue satisfies λ = 0, 0 < λ < 1, λ = 1 or λ > 1. Back to Example 32: Let us classify the 2-cycle (0, 1). F 2 (x) = F (F (x)) = (x 2 1) 2 1 = x 4 2x 2. (F 2 ) (x) = 4x(x 2 1) is such that (F 2 ) (0) = (F 2 ) ( 1) = 0, so the cycle is indeed superattracting and attracts all seeds in the set ( x +, x + ) except at ±x. Remark 14. We may notice that F 2 above is a polynomial of degree 4, so F (x) x = 0 has two other roots, i.e. F 2 has two other fixed points. Could we have forgotten another 2-cycle along the way? No, the two remaining points are the first two fixed points of F itself x + and x (if F (x) = x, then certainly F 2 (x) = F (F (x)) = F (x) = x), which can be seen as two 1-cycles. The quadratic family. Examples 31 and 32 seem to come from very similar functions (x 2 and x 2 1), yet the dynamical behaviors exhibited show quite different features. Considering now the one-parameter family of functions F c (x) x 2 + c for c R, we will see that varying continuously the parameter c can lead to drasting changes in dynamical behavior as c passes certain values. As in both examples considered above, let us start by searching the fixed points of F c. F c (x) = x is equivalent to, after completing the square, (x 1 2 )2 + c 1 4 = 0. A first observation, then, is that if c > 1 4, F c has no fixed point. Graphically, the parabola lies completely above the curve {y = x}, and graphical analysis shows that all orbits have limit +. The natural way of evolving c is then to turn it down starting from above 1 4 and see what happens to the dynamics. When c = 1 4, the parabola is tangent to the curve {y = x} at the point ( 1 2, 1 2 ), and x = 1 2 is a fixed point with F 1 ( 1 2 ) = 1, so the point is neutral. By graphical analysis, we arrive at the 4 following conclusion: if x > 1 2, the orbit of x has limit +. if x 1 2, the orbit of x has limit 1 2. When c < 1 4, F c has now two fixed points x ± = 1 2 ± 1 4 c with F c(x + ) = c > 1 so x + is repelling for all c < 1 4. On the other hand, F c(x ) = 1 1 4c, so x can still exhibit all types of stability depending on the value of c. It is always true that F c(x ) < 1 but forcing F c(x ) > 1 entails c > This means that in the regime 4 < c < 1 4, x is either attracting or

59 59 superattracting. Graphical analysis now yields the following conclusions: If x > x +, the orbit of x has limit +. If x = ±x +, the orbit of x has limit x +. If x x, the orbit of x has limit x. Figure 6: The saddle-node bifurcation. From left to right, dynamical pictures for values of c (0.4, 1/4, 0.5). The dynamical transition that occurs as c passes the threshold 1 4 is called a saddle-node bifurcation. Saddle because in the case c = 1 4, the point x = 1 2 can be seen as a saddle point, whose attracting or repelling properties depend on what side of the point you are looking. Node because for c slightly less than 1 4, the pair x +, x, respectively repelling and attracting, is called in other contexts a source (out of which orbits leave) and a sink (into which orbits converge) and the pair is referred to as a node.

60 60 Exercises for lecture 20: 1. For each of the following discrete-time dynamical systems x n+1 = F (x n ), find all fixed points and determine their stability (pictures can help!): (a) F (x) = x 3 2 (b) F (x) = x 2 + 2x 1 (c) F (x) = e x Carry out a full analysis of the system F (x) = (x 1) 2 + 2, involving fixed points and their stability, graphical analysis, potential 2-cycles and their stability. 3. Using the chain rule, prove that for a k-cycle (x 0, x 1,..., x k 1 ), we have (F k ) (x 0 ) = (F k ) (x 1 ) = = (F k ) (x k 1 ) = (Hint: you may first start with a 2-cycle). k 1 j=0 F (x j ). Material used for this lecture: [DK] Dynamics of simple maps, R. Devaney.

61 61 Lecture 21 - Study of the quadratic family, continued. The period-doubling bifurcation. So far, we have seen in the regime 1 4 > c > 3 4 that F c has two fixed points x ± = 1 2 ± 1 4 c, that x + is repelling and x is attracting. At this point, we should also add an important observation that for 1 4 > c 2, the interval [ x +, x + ] is stable under F c, that is to say F c ([ x +, x + ]) [ x +, x + ]. In particular, we see that if a seed is in this interval, its entire orbit must remain inside of it. As c decreases through the value 3 4, F (x ) becomes greater than 1, then x becomes a repelling fixed point. Since the [ x +, x + ] is still stable, what type of dynamics is happening there? The answer is that the point x, when becoming repelling, gives birth to an attracting 2-cycle in the process, which will attract all orbits that lie inside ( x +, x + ) (except the fixed orbit x ). We see this by studying Fc 2, whose number of real roots switches from 2 when c > 3 4 to 4 when c < 3 4. Indeed, Using the fact that fixed points of F c are fixed points of Fc 2, we can factor F c (x) x out of the polynomial Fc 2 (x) x and using the method of undetermined coefficients, establish that Fc 2 (x) x = (x 2 + c) 2 + c x = (F c (x) x)(x 2 + x + c + 1) ( = (x x + )(x x ) (x )2 + c + 3 ). 4 We see that if c + 3 4, the last polynomial has no real roots and when c , then we obtain two new fixed points, say x ± = 1 2 c ± 3 4, these are the ones forming an attracting 2-cycle. This transition from an attracting fixed point to an attracting two-cycle is called a period-doubling bifurcation. Figure 7: The period-doubling bifurcation. Left: c = 0.5 > 3 3 4, right: c = 1 < 4.

62 62 The period-doubling route to chaos. Before c reaches 2, this is certainly not the only bifurcation that occurs. In fact, period-doubling bifurcations will occur several times and lead to quite complicated-looking orbit. Analyzing this in detail would be a lost cause and would require computing roots of polynomials of arbitrarily high order. However, one graphical way of seeing what happens is, for several values of c between 1 4 and 2, plot on the y-axis some of the iterates {Fc n (0)} for 100 n 150. We are dropping the first 100 terms assuming that the transitory regime of the orbit will be gone by then, and that the orbit of 0 will have reached either convergence, or a periodic orbit. The plot obtained can be seen Figure 8. Note that the choice of initial seed inside [ x +, x + ] does not matter as in this regime, almost all orbits trapped in this interval have the same long-term behavior. Figure 8: The bifurcation diagram of the orbits of x 0 = 0 under the dynamics F c (x) = x 2 + c. The Cantor middle-third set. Before finishing our excursion through the behavior of the quadratic family, we must study one of the first fractal sets: the Cantor set C. We construct C as follows. Start from the segment [0, 1] and remove the middle-third open interval ( 1 3, 2 3 ). Out of the remaining set [0, 1 3 ] [ 2 3, 1], remove the middle-third open interval of each remaining interval, that is, remove ( 1 9, 2 9 ) and ( 7 9, 8 9 ). Repeat this process over and over ab infinitum, and define the Cantor set to be ( 1 C = [0, 1] 3, 2 ) ( 1 3 9, 2 ) ( 7 9 9, 8 ) ( , 2 ) ( , 8 ) ( , 20 ) ( , 26 ) At the end of this process, how much length have we removed (in other words, what is the Lebesgue measure of C?). The answer is = 1 ( 2 ) n 3 n=0 3 = = 1! So the Cantor set seems to have length 0. On the other hand, it can be shown that it still contains as many points as the entire segment [0, 1]! Among other properties, here are some of the properties of this strange animal: C has empty interior, it contains no open interval (note that, otherwise, it would have nonzero length).

63 63 Every point of C is an accumulation point of C. These first two properties make C a so-called perfect set. C is uncountably infinite, in particular there is a bijection from C to [0, 1]. C is self-similar, i.e. it is the union of two copies of itself! This is what makes C a fractal set: the left half of C looks like C, so does the left half of the left half, etc... The regime c < 2. Chaotic behavior. We would now like to study what happens when c < 2, for which the definition of Cantor set above comes in handy. When c < 2, something new and dramatic happens: the interval [ x +, x + ] is no longer stable under F because the bottom of the parabola is poking through the box [ x +, x + ] [ x +, x + ]. This will cause some orbits that initially started inside [ x +, x + ] to actually converge to +. The question is now, do all orbits eventually converge to +? The method to study that is to follow backwards the orbits that go to + and cover all possible scenarios. Graphically, we can find out by successive elimination of all orbits converging to + that the seeds of orbits remaining in [ x +, x + ] form at most a set that is homeomorphic to the Cantor set above. Call this set Λ. At this point, we will merely state facts about what actually happens, though proving anything would be beyond the scope of this introduction. There indeed exists a set Λ [ x +, x + ] that is homeomorphic to a Cantor set, and such that F c (Λ) Λ, so we can consider the dynamical system restricted to Λ. The dynamics on Λ is called chaotic in the sense of the definition below, which requires putting an appropriate topology on Λ (the topology of open intervals is now inappropriate since we know that Λ does not contain an open interval and looks like a collection of discrete points). This is done by defining a certain metric function d(x, y) on Λ, which in turn allows to reintroduces the notion of open set and neighborhood (i.e. neighborhoods would be spanned by sets of the form D δ (x) = {y Λ, d(x, y) < δ}). For a special choice of such a distance function, we can show that F c : Λ Λ is chaotic in the sense of the definition below. Definition 22. Let (Λ, d) a metric space with distance function d(x, y) and F : Λ Λ. F is called chaotic if the following conditions are satisfied (i) F is topologically transitive, this means that for every U, V open sets in Λ, there exists k such that F k (U) V. (ii) F has sensitive dependence to initial conditions, this means that there exists δ > 0 such that for every x Λ and every neighborhood V of x there exists y V and n N such that d(f n (x), F n (y)) δ. 8 (iii) Periodic points of F are dense in Λ. 8 Equivalently, there exists δ > 0 such that for every x Λ and every ε > 0, there exists y Λ and n N such that d(x, y) < ε and d(f n (x), F n (y)) > δ.

64 64 Point (i) means that no matter the sets U, V, there will always be a seed in U whose orbit will visit V. Point (ii) means that there will always be points arbitrarily close to one another whose orbits will be separated by an absolute distance. As Edward Lorentz describes it, Chaos is when the present determines the future, but the approximate present does not approximately determine the future. It is essentially a description of lack of predictability, despite the fact that the system is completely deterministic.

65 65 Lecture 22 - Complex dynamics. Dynamics under the map z z 2 Now that an example of one-dimensional dynamics has showed us a variety of dynamical behaviors, we are ready to explore two-dimensional dynamics. The function to be iterated will now be a function of a complex variable. In particular, we will use a rational function F (z) = P (z) Q(z), where P and Q are polynomials. Then F can be extended into an analytic function F : C C, and we must study the dynamical system z 0 C, z n+1 = F (z n ), n N. As in the one-dimensional case, we define z to be a k-periodic point of F if F k (z ) = z and for every 1 j < k, F j (z ) z. The orbit of a k-periodic point is periodic with period k, i.e. it is of the form z, z 1,..., z k 1, z, z 1,..., z k 1, etc... and it is called a k-cycle. Of course, the case k = 1 covers the case of fixed points of F. For a k-periodic point z, we call its multiplier λ = (F k ) (z ), whose modulus allows us to tell whether the point z is superattracting (λ = 0), attracting ( λ < 1), neutral ( λ = 1) or repelling ( λ > 1). The point can also be a fixed point, in which case the multiplier to consider is λ = 1 F ( ). Finally, for a fixed point z of F, we define the basin of attraction of z as the set of seeds in C whose orbits converge to z, that is, A F (z ) = {z C : lim n F n (z) = z }. Remark 15. Because when a limit is unique whenever it exists, the basins of attractions of two distinct fixed points cannot overlap. So each basin of attraction covers a certain part of C. The question is, do they cover the whole sphere? If not, what happens to the dynamics outside the basins of attraction? The answer is, chaotic behavior. Let us now study an example in detail. An example of a chaotic function. Let us study the dynamical system associated with the function F (z) = z 2. In this case, we can actually compute the iterates explicitely and arrive at F n (z) = z 2n. We first look for the fixed points of F, which after solving F (z) = z yields z {0, 1}. We also need to consider and we notice also that F ( ) = so the fixed points are z {0, 1, }. The multipliers of each of these points are (F (0), F 1 (1), F ( )) = (0, 2, 0), respectively, so 0 and are superattracting while 1 is repelling. From these basic observations, we realize that 0 and will be competing to attract orbits. The interface between these basins of attraction will be made of points whose orbits should not be expected to converge anywhere and will trigger chaotic dynamical behavior. Further straightforward observations lead to the following partitioning of the Riemann sphere into three sets, each of which is stable by F : C = U 1 U 2 U 3, where U 1 = D, U 2 = D and

66 66 U 3 = C D and for 1 j 3, F (U j ) U j. This is because due to the form of F, we have that z < 1 implies F (z) < 1, z = 1 implies F (z) = 1 and z > 1 implies F (z) > 1. Moreover, if z 1 < 1, then lim n z 2n = 0 and if z > 1, lim n z 2n =. So we have completely characterized the basins of attraction of 0 and, that is A F (0) = D and A F ( ) = C D. It remains to study the behavior of F on the unit circle D. Theorem 32. F (z) = z 2 is chaotic as a mapping from D to itself. Note that when discussing about chaos in the previous lecture, we had trouble understanding what it meant to put an ad hoc topology on Cantor sets. Here we can work with the usual distance on the unit circle, i.e. for two points on the unit circle x = e iα and y = e iβ, we will denote d(x, y) = mod (α β, 2π). For this distance, the basic open sets are open arc circles of the form {e iθ, θ (α ε, α + ε)}, this particular one being a neighborhood of e iα. Proof of Theorem 32. We need to show that F is (i) topologically transtive, (ii) has sensitive dependence on initial conditions and (iii) its periodic points are dense in D. Proof of (i). We need to show that for U, V two open sets, there exists k > 0 such that F k (U) V. Pick U an open set, then it contains a basic open set of the form U 0 = {e iθ, θ (α ε, α + ε)}. For any k > 0, we have F k (U 0 ) = {e i2kθ, θ (α ε, α + ε)} = {e iθ, θ (2 k α 2 k ε, 2 k α + 2 k ε)}. So once k is such that 2 k ε > π, F k (U 0 ) contains the whole unit circle, so no matter what V is, F k (U) V. F is therefore topologically transitive. Proof of (ii). We need to prove that there exists δ > 0 such that for every x D, for every ε > 0, there is y and k > 0 such that d(x, y) < ε and d(f k (x), F k (y)) > δ. Let us show that δ = π 2 does the trick. Pick x = e iα and for ε > 0 small, y = e i(α+ε). As in point (i), d(f k (x), F k (y)) = 2 k ε so for k large enough, we will obtain π > 2 k ε > π 2 (find that k), so that the claim is proved. Proof of (iii). The periodic points of F are any root of F k (z) z for any k > 0. In particular, we solve 0 = F k (z) z = z 2k z = z(z 2k 1 1), so we find that the k-periodic points of F are the 2 k 1-th roots of unity, which we know form a regular 2 k 1-gon on the unit circle. As k increases, the number of vertices of this polygon becomes larger and larger and in the end, all the vertices of these polygons form a dense subset of the unit circle. Hence the proof. Additionally, it is instructive to look at the 2-cycles and 3-cycles of F. The 2-periodic points of F on the unit circle are the third roots of unity (1, e 2iπ 3, e 2iπ 3 ). Since z = 1 is a fixed point of its own, the remaining two roots for a 2-cycle (e 2iπ 3, e 2iπ 3 ). The 3-periodic points of F on the unit circle are seventh roots of unity e 2ikπ 7 for k = Again, z = 1 will not give a 3-cycle because we know it s a 1-cycle. The 6 remaining points will form two 3-cycles as shown in Figure 9.

67 67 Figure 9: Two-cycles (left) and three-cycles (right) of the mapping F (z) = z 2. Exercises for Lecture 22: 1. Under the mapping F (z) = z 2, compute and plot the 6 first terms of the orbits for the following seeds: 0.5i, 0.5, 2, 2e i π Find all 4-periodic points of F (z) = z 2 and draw the 4-cycles using a regular 15-gon as in Figure We have seen in class that the point z = 1 is a repelling fixed point of z z 2. Nevertheless some orbits eventually station at z = 1. Describe the set of seeds whose orbits eventually station at z = 1. [Hint: start solving F (z) = 1, then F 2 (z) = 1, etc... ]

68 68 Lecture 23 - Global conjugacies and application to the study of Newton s algorithm for quadratic polynomials Recall that we are studying the dynamics associated with rational maps of the form F (z) = P (z) Q(z) with P, Q polynomials, i.e. functions who are analytic from C to itself. A way of classifying these maps by means of their dynamical behavior is via the concept of global conjugacy: Definition 23. Given two rational maps F, G : C C and a linear fractional transformation φ, we say that F and G are globally conjugate via the conjugacy φ if F φ = φ G, i.e. F (φ(z)) = φ(g(z)), z C. This implies the relation G = φ 1 F φ or equivalently F = φ G φ 1. These relations can be illustrated by saying that the graph below is commutative (e.g. one can map the lop-left C to the top-right C by either mapping via F, or by mapping via φ then G then φ 1 ). C φ C G F C φ C Theorem 33. Being globally conjugate is an equivalence relation among rational maps. Proof. Clearly this relation is reflexive since for any F, we have F = Id F Id 1. This relation is symmetric since if we have F = φ G φ 1, then G = φ 1 F φ, where φ 1 is an LFT whenever φ is. Finally this relation is transitive since if we have F = φ 1 G φ 1 1 and G = φ 2 H φ 1 2 for some LFT s φ 1, φ 2, then F is conjugate to H via φ 1 φ 2 since we have the relation This completes the proof. F = φ 1 φ 2 H φ 1 2 φ 1 1 = φ 1 φ 2 H (φ 1 φ 2 ) 1. Now let us state a few properties justifying why two globally conjugate maps exhibit the same dynamical behavior. First notice that if G = φ 1 F φ, then a simple calculation shows that so that one can prove by induction that G 2 = φ 1 F φ φ 1 F φ = φ 1 F 2 φ, }{{} Id G k = φ 1 F k φ, k N. Theorem 34. Let F, G be conjugate via the relation G = φ 1 F φ. Then the following properties hold true:

69 69 (i) For any k 1, z is a k-periodic point of G if and only if φ(z ) is a k-periodic point of F. Moreover, their multipliers are that the same, i.e. (G k ) (z ) = (F k ) (φ(z )) 9. (ii) For z an attracting fixed point of G, then by virtue of (i), φ(z ) is an attracting fixed point of F, and then φ(a G (z )) = A F (φ(z )). (iii) The set where G is chaotic is mapped bijectively by φ onto the set where F is chaotic. Proof. We only treat points (i) and (ii). First notice that if G = φ 1 F φ, then for any k 1, G k = φ 1 F k φ, so that G k is globally conjugate to F k via the same conjugacy φ. Equivalently, we have φ G k = F k φ. Proof of (i): Fix any k 1. If z = G k (z ), then φ(z ) = φ G k (z ) = F k φ(z ) = F k (φ(z )), hence φ(z ) is a k-periodic point of F, the converse is also true by symmetry of the conjugacy relation. Differentiating the relation φ G k (z) = F k φ(z), we obtain φ (G k (z))(g k ) (z) = (F k ) (φ(z))φ (z). Upon evaluating this equality at z, using that z = G k (z ) and the fact that φ(z ) 0 because φ is conformal, we arrive at (G k ) (z ) = (F k ) (φ(z )), hence the multipliers are equal. Proof of (ii): Let z A G (z ). This means that lim n G n (z) = z. Applying φ, we obtain φ(z ) = φ( lim n Gn (z)) = lim n φ Gn (z) = lim n F n (φ(z)). So if z A G (z ), then φ(z) = A F (φ(z )). The converse is also true by symmmetry of the conjugacy relation, so that the result holds. Classification of Möbius transformations. In the language of conjugacies, the classification of LFTs established in lecture 9 can be reformulated as follows: If F is a Möbius transformation with a single fixed point (i.e., the parabolic case), then there exists b C such that F is globally conjugate to the mapping z z + b. If F is a Möbius transformation with two fixed points, then there exists m C {0} such that F is globally conjugate to the mapping z mz. In this case, F is elliptic when m = 1, hyperbolic when m (0, + ), loxodromic otherwise. 9 This implies that their types (attracting, superattracting, neutral, repelling) are the same.

70 70 Newton s algorithm for quadratic polynomials. Let us now see how we can use conjugacies in order to study the dynamics of the Newton algorithm applied to quadratic polynomials, restricting to the case of two distinct roots. Consider the function f(z) = (z α)(z β) with α β. The corresponding Newton map is F (z) = z f(z) f (z) 10. After simplification, we find F (z) = z2 αβ 2z (α+β). We will show the following. Proposition 35. Let f(z) = (z α)(z β) a quadratic polynomial with two distinct roots α β. Then the Newton map of f defined by F (z) = z f(z) f (z) is globally conjugate to the mapping z z2. Proof. Define the LFT φ(z) = βz+α z+1, then a direct calculation shows that F φ(z) = φ(z 2 ) = φ G(z), G(z) = z 2. (8) Remark 16. The construction of φ 1 (z) = z α z β is motivated by the fact that if φ were such a conjugacy, by virtue of Theorem 34, φ 1 would map fixed points of F to fixed points of G and the multipliers of these fixed points would have to be the same. This is indeed satisfied, as G(z) = z 2 has fixed points (0, 1, ) with multipliers (0, 2, 0). F has fixed points (α, β, ) with multipliers (0, 0, 2). Therefore, φ 1 must map (α, β, ) to (0,, 1), which is exactly what it does. Combining the analysis of the map G(z) = z 2 done in the previous lecture, Proposition 35 and Theorem 34, we may draw the following conclusions: for F (z) = z2 αβ 2z (α+β) the Newton map of the quadratic polynomial f(z) = (z α)(z β), we have that A F (α) = A F (φ(0)) = φ(a G (0)) = φ(d). A F (β) = A F (φ( )) = φ(a G ( )) = φ(c D). F is chaotic on φ( D). A closer look at the mapping φ(z) = βz+α z+1 tells us that φ( D) is the median line between α and β, call it L { } (note that φ(1) =, and since φ is an LFT, it must map the circle D to a circle in C passing through, that is to say, a straight line). The points on this line are precisely where F is chaotic, i.e. where the Newton algorithm fails to converge to a root of f. A heuristic interpretation of this would be: if a seed is equidistant to both roots α and β, an argument of symmetry tells us that there is a priori no reason for this seed to be attracted more by one root or the other notice that this expression would not change if we had multiplied f by a constant, so there is no loss of generality in assuming this form for f 11 We will see that this heuristic argument completely falls through in the case of a cubic polynomial.

71 71 Our conclusion on the Newton algorithm for a quadratic polynomial with two distinct roots is that, under the dynamical system F (z) = z f(z) f (z), any seed has its orbit converge to whatever root of f is is closer to. If it is equidistant to both roots, then the algorithm fails to converge and the dynamics is, in fact, chaotic.

72 72 Exercises for Lecture 23: 1. Prove (8). 2. Study the Newton algorithm applied to a generic quadratic polynomial with a double root, of the form f(z) = (z α) Let f 1 (z) = (z α 1 )(z β 1 ) and f 2 (z) = (z α 2 )(z β 2 ), both with distinct roots and with corresponding Newton maps F 1 and F 2. Find an explicit conjugacy φ such that F 1 = φ 1 F 2 φ. [Hint: first explain why this conjugacy should map the triple (α 1, β 1, ) to (α 2, β 2, ).] 4. Let f(z) = z and F (z) = 1 2 (z 1 z ) its Newton map. (a) Find a global conjugacy between F and G(z) = z 2. (b) Using this conjugacy and what we know about periodic points of G, find a way to describe all periodic points of F. 5. Prove that the quadratic map F (z) = az 2 + bz + d is globally conjugate to the quadratic map G(z) = z 2 + c only if their coefficients satisfy the relation c = ad + b 2 b2 4. [Hint: a necessary condition for F and G to be conjugate is that the multipliers of their fixed points must match.]

73 73 Lecture 24 - Invariant sets - Julia and Fatou sets Newton s algorithm for cubic polynomials. We ve seen above that for any quadratic polynomial with distinct roots, the dynamical behaviors are quite similar to one another as they are all, in fact, pairwise conjugate. The case of cubic polynomials, involving now the interplay between three basins of attraction instead of two, not only leads to a higher diversity of behavior, but also higher complexity of sets. A first observation is that not all cubic polynomials are pairwise conjugate, however an elementary family to study is the following family p ρ (z) = z(z 1)(z ρ), with roots (0, 1, ρ) with ρ U = {Im (z) > 0} D D 1 (1). Theorem 36. For any polynomial with three distinct roots f(z) = (z α)(z β)(z γ), there exists a linear transformation T (z) = az + b and ρ U such that the Newton map F is globally conjugate to the Newton map of p ρ via T. Because ρ is now a free parameter, that leaves a lot of room for different dynamical behaviors! Now one may wonder, when picking a cubic polynomial, thereby creating three superattracting fixed points in the Newton dynamics, is it still true that a seed will converge to the roots that it is closest to? In the case of the polynomial f(z) = z 3 1 should we draw the three medians in between each pair of roots of unity and determine the basins of attraction in this way? The answer turns out to be much more complicated, as one may see on Figure 10 (also see Fig. 11 for another example of cubic). As a foretaste, the main constraint leading to such complicated behavior is the following result, of which we will provide a proof later. Theorem 37. If F is a rational function and z 1, z 2 are two attracting fixed points of F, then necessarily A f (z 1 ) = A F (z 2 ). In the case of f(z) = z 3 1, we thus require that A f (1) = A f (e 2πi 3 ) = A f (e 2πi 3 )! We will now introduce notions that will help us prove Theorem 37. Topological preliminaries Here and below, we consider F : C C analytic (i.e. a rational function) and U a subset of C. For U C we denote by U C = C U the complement of U. Definition 24. We say that (i) U is invariant (under F ) if F (U) U. (ii) U is completely invariant (under F ) if F (U) U and F 1 (U) U.

74 74 Figure 10: Partitioning of the plane into the basins of attractions of the roots of z 3 1. at different levels of zooming. Figure 11: Partitioning of the plane into the basins of attractions of the roots of z(z 1)(z + 1), at different levels of zooming.

75 75 Example 33. If F (z) = z 2, then D is completely invariant, and {1} is invariant but not completely invariant, since F (1) = 1 but F 1 (1) = { 1, 1}. Lemma 38. F 1 (U) U if and only if F (U C ) U C. As a consequence, U is completely invariant if and only if U C is completely invariant. Proof. ( ) Suppose F (U C ) U C. Let y F 1 (x) where x U, in particular f(y) = x. If y U C, then by the assumption, x = f(y) U C which is a contraction, therefore y U. Hence F 1 (U) U. ( ) Suppose F 1 (U) U. Let x U C. If F (x) U, we have x F 1 (F (x)) F 1 (U) U which is a contraction, hence F (x) U C. Hence F (U C ) U C. Lemma 39. If F is an open mapping (i.e. it maps open sets to open sets) and U is invariant, then U o, the interior of U, is invariant. Proof. If y U o there exists an open set V such that y V U o. neighborhood of F (y) included in U, that is, F (y) U o. Then F (V ) is an open Lemma 40. If F is a continuous open mapping 12, and U is a completely invariant open set, then so is U. Proof. Since U is open then U U =. Let x U. There exists x n U such that x n x. Since F (U) U, the sequence F (x n ) is in U and therefore its limit is in U = U U. Since U is completely invariant and x / U, F (x) / U so F (x) U. Therefore F ( U) U. Let us now prove that F 1 ( U) U by showing that F (( U) C ) ( U) C. Indeed, ( U) C is the disjoint union of U (which is F -invariant) and (U C ) o, which is F -invariant since U C itself is and using the previous lemma. We finish the preliminaries by stating without proof another theorem of Montel s theorem, establishing the property of a sequence of meromorphic functions to have a normally convergent subsequence, under much weaker assumptions. Theorem 41 (Montel s theorem (hard version)). A sequence of meromorphic functions f n : U C which omits three values in C (in the sense that the set C n N f n(u) contains at least 3 distinct points) has a normally convergent subsequence. Remark 17. Note that if a sequence is uniformly bounded (the condition for the first Montel s theorem), then it clearly omits at least three points (in fact C n N f n(u) is an unbounded set!). 12 The Open Mapping Theorem says that this is always satisfied when F is analytic and non-constant.

76 76 Julia and Fatou sets. We are now ready to introduce the concept of Fatou and Julia sets, named after Gaston Julia and Pierre Fatou, who initiated the study of iterated rational functions in the 1920 s. Definition 25 (Julia and Fatou sets). Given F : C C a rational function, we define the Fatou (or stable) set of F (denoted F(F )) and Julia set of F (denoted J (F )), as follows: 1. z 0 F(F ) if there exists ρ > 0 such that the sequence of iterates {F n } n has a normally convergent subsequence on D ρ (z 0 ). 2. J (F ) is the complement in C of F(F ). By construction, the Fatou set is open, then the Julia set is closed. The Fatou set is where the dynamics is well-behaved, whereas the Julia set is where the dynamics is chaotic. Proposition 42. The Fatou set F(F ) is completely invariant under F. As a consequence, so is the Julia set J (F ). Proof. We must show for every z 0 C, z 0 F(F ) if and only if f(z 0 ) F(F ). ( = ) Suppose f(z 0 ) F(F ), then there exists U an open neighborhood of f(z 0 ) and a sequence {n k } k of integers where the sequence {F n k} k converges normally. Then f 1 (U) is an open neighborhood of z 0 where the sequence {F n k+1 } k converges normally, hence z 0 F(F ). ( = ) Similar proof, except that one must use the fact that F is an open mapping to say that it maps an open neighborhood of z 0 into an open neighborhood of f(z 0 ). Some properties of the Julia set. For an integer n, we denote F n (w) the set F 1 (F 1... (F 1 (w))), where the preimage is taken n times. In what follows, we denote F a rational function and J its Julia set. Proposition 43. For w J, the set n N F n (w) is dense in J. Proof. Let w J. We want to show that for every z J and every neighborhood V z, there exists k such that F k (w) V. By the contraposition of Montel s theorem above, {F n } n is not normal on V so C n F n (V ) is at most two points and it can be shown by exhaustion of cases that these exceptional points belong to the Fatou set, so w is not one of them. Therefore, w n F n (V ), i.e. there exists k such that w F k (V ). That means that F k (w) V, hence the result. Example 34. For F (z) = z 2, J (F ) = D, then we have seen in a previous exercise that n N F n (1) is the union over n N of all 2 n -th roots of 1. This set is clearly dense in the unit circle. Using Proposition 43, we can establish the following Proposition 44. J has no proper subset that is both closed and completely invariant.

77 77 Proof. Let K be a subset of J satisfying both properties, and let w K. Then since K is completely invariant, n N F n (w) K. Taking topological closure and using the previous proposition, we obtain that J n N F n (w) K = K, hence K = J. Application to Newton s algorithm for cubic polynomials We can now return to the proof of theorem 37. As mentioned in lecture 22, if f(z) = (z w 1 )(z w 2 )(z w 3 ) is a cubic polynomial and F is its Newton map, F has three superattracting fixed points at w 1, w 2, w 3, each of which has an open basin of attraction A F (w k ). Each basin of attraction is completely invariant, so by the preliminaries, A F (w k ) is completely invariant as well. One can also show that this set is closed, and a quick moment s thought makes one think that it could not be part of the Fatou set (otherwise, that basin of attraction could be extended a little further). Therefore for each k = 1, 2, 3, A F (w k ) is a closed, completely invariant subset of J, which by virtue of Proposition 44, implies that A F (w 1 ) = A F (w 2 ) = A F (w 3 ) = J (F ). Enforcing this condition clearly leads to the complicated intertwining that is occuring on the pictures (see Figure 12) showing basins of attraction: any neighborhood of any point in J (F ) must contains points which belong to each of the basins of attraction! To generalize further, you may imaging that Newton s dynamics for polynomials of degree d 3 may be such that the Julia set must be the common boundary to d disjoint basins of attractions.

78 Figure 12: More examples of basins of attractions for the Newton dynamics of various sample cubic polynomials. The roots are in yellow. 78

79 79 Exercises for lecture 24: 1. Prove Theorem 36. [Hint: the three roots form a triangle. Pick the two roots forming the longest edge and map them to 0 and 1 using a linear transformation.]

80 80 Lecture 25 - Zoology of Fatou sets. Representation of Julia sets. Discussion about the Fatou set. Here, we will merely state facts about the types of behavior one can expect on the Fatou set. We have seen an example before, which was the basin of attraction of a fixed points. Let Ω a connected component of the Fatou set. Since the family {F n } has normally convergent subsequences there (call accumulation function its limit), one may ask the question: how many accumulation functions does the sequence {F n } n have? If it is a finite number, we may expect to be in the basin of attraction of an attracting/superattracting or neutral periodic cycle. In the latter case, such connected components are referred to as parabolic cycles. For example, it can be shown that a neighborhood of a neutral fixed point is split into an even number of sectors, where orbits alternatively converge or diverge away from that fixed point. Finally, if the number of accumulation functions of the sequence {F n } n is infinite (think for instance of the sequence of iterates of the map F (z) = e 2iπα z with α irrational), depending on the connectedness of Ω, it may lead to a so-called Siegel disk or a Herman ring. The interested reader may find more detail on this topic in [DK], Article Julia sets by Linda Keen. Quasi self-similarity of Julia sets. The fractal quality of Julia sets is formalized by the concept of quasi-self-similarity. Selfsimilarity occurs when a set is a dilated, isometric copy of one of its proper subsets, see e.g. the Von Koch curve or the Cantor set. In our case, Julia sets are rarely self-similar because the smaller copies are slightly distorted, though this distortion remains bounded in some sense. One may thus think of the concept of quasi-self-similarity as a self-similarity up to small distortions. Definition 26. φ : U V is a K-quasi-isometry (K 1) if for every x, y U, we have 1 x y φ(x) φ(y) K x y. K If K = 1, then φ(x) φ(y) = x y and φ is an isometry (e.g. in Euclidean space, a composition of rotation and translation). Theorem 45 (Sullivan). If F is a rational, expanding map, then the Julia set of F is such that there exists K 1 and r 0 > 0 such that for every z J and every 0 < r < r 0, the set J D r (z), dilated by a factor 1 r, is K-quasi-isometric to the whole of J. Numerical representation of Julia sets sets. Method 1: Inverse iteration method Let us present two methods in order to visualize Julia The first method is based on Proposition 43, i.e. that for any w J, the set n N P n c (w) is dense in J. The method therefore consists in (i) finding a point w 0 in J and (ii) computing successively the preimages P n c (w 0 ).

81 81 to r.), illustrating quasi-self- Figure 13: Examples of zooms into a Julia set (zoom-ins from l. similarity Example 35. Fix c C and define P c (z) = z 2 +c. For this function, the two steps described above ( go as follows: Step (i): We know that P c (z) = z 2 + c has fixed points z ± = 1 2 ± 1 4 c) and that z + is always repelling, therefore z + J. Step (ii): Computing preimages of a given w consists in solving for z the equation z 2 +c = w, which obviously has two solutions z = ± w c, and these are the two preimages of w. So by induction, we can establish that Pc (n+1) (w 0 ) = {± w c, w Pc n (w 0 )}. We can see that Pc n (w 0 ) has exactly 2 n distinct elements. Method 2: Boundary scanning method The main drawback of the previous method is that it tries to capture the Julia set spot on, whose chaotic nature does not pair well with the finite-accurary computations of computers. The second method, is based on the observation that for a function of the form P c (z) = z 2 + c, is a superattracting point, so by lecture 22, it has a basin of attraction, and moreover, J (P c ) = A Pc ( ). A quick estimate also shows that if R = max(2, c ), then J D R (0) because { z > R } A Pc ( ). The method then goes as follows: Fix a large integer N max (say 500). (i) Discretize a fine enough grid of the square [ R, R] [ R, R]. (ii) For each gridpoint z, compute its orbit for N max steps and set n(z) to be either the smallest integer where P n c (z) > R, or N max if the orbit never exitted D 2 (0). (iii) Display n(z). Instead of focusing on the boundary itself, we focus on visualizing A Pc ( ) as best we can, by computing at each point of a given grid in how many steps the orbit of the point arrives in A Pc ( ). n(z) is sometimes referred to as the escape rate. One may notice that in some cases, a whole region of the plane is colored with the value N max, this is because in some cases, the Fatou set of P c is made of two components, one of which is bounded by the Julia set. We sometimes refer to this region as the filled-in Julia set. We will see that the condition of whether there exists another bounded component of the Fatou set of P c is the very criterion which defines the Mandelbrot set.

82 Figure 14: Examples of Julia sets using both visualization methods on the iteration of polynomials of the form P c (z) = z 2 + c. 82

83 83 Figure 14 shows examples of Julia sets for four different values of c. We see that the first method, though much less expensive, lacks accuracy because the function that is iterated over is z instead of z (the first method relies on inverse orbits while the second relies on forward orbits). This explains why in the outcome of method 1, some points lie outside the main cloud. Except in the bottom-left example, this is NOT to be expected, as one can show that the Julia set is in fact connected. Another advantage of Method 2 over Method 1 is that, for polynomials of degree > 2, it becomes tricky to compute preimages by polynomials of high degrees. This can still be done with relative ease on the function F (z) = z z3 1 as seen on Fig. 15 (the Newton map of z z 3 1), though 3z 2 in most cases, it seems much more convenient to use the boundary scanning method, which is easy to implement for polynomials of arbitrary order... Figure 15: Top: Julia set of Newton map of f(z) = z 3 1 via inverse iterates. Bottom: same Julia set obtained via the boundary scanning method.

84 84 Exercises for lecture 25: 1. Describe the steps to achieve the inverse iterates method constructing the Julia set of the map F (z) = z z3 1 (Newton map of f(z) = z 3 1). [Hint: upon solving for z the equation 3z 2 F (z) = w, notice that z = 1 z solves a reduced cubic equation (z ) 3 + pz + q = 0, for which the roots have explicit expressions which you may find online.] 2. (optional) Complete the IIMcubic.m file to implement the algorithm you find. It will work once you obtain the top plot of Fig. 15. Material used for this lecture: Article Julia sets, by Linda Keen in [DK].

85 85 Lecture 26 - Critical points and the Mandelbrot set It is now time to stress the importance of the orbits of critical points in complex dynamics. Theorem 46 (Fatou, 1905). Every attracting cycle for a rational function attracts at least one critical point. This means that if F = P Q is a rational function with P, Q polynomials of degree at most d, the equation F = 0 consists in finding the roots of P Q P Q, a polynomial of degree at most 2d 1, so F has at most 2d 1 critical points and can only have at most 2d 1 attracting cycles. In the case of the family P c (z) = z 2 + c, we have P c(z) = 2z regardless of c, and z = 0 is the only critical point, so P c has at most one attracting cycle. Another important theorem is the following. For F rational, denote K F the set of seeds with bounded orbit, and Ω F the set of critical points of F. Theorem 47 (Fatou, Julia). If F is a polynomial, then we have: (i) Ω F K F if and only if J F is connected. (ii) if Ω F K F =, then J F is a Cantor set. Since Ω F = {0} when F = P c, these are the only cases to consider. The c-plane is then split into two sets where either 0 K Pc or 0 / K Pc. Definition 27. The Mandelbrot set M is the set of c in the parameter plane such that case (i) in Theorem 47 occurs. Remark 18. The Mandelbrot set is a set in the c-plane, or the parameter plane, whereas the Julia set is a set of the dynamical plane z. Each value of c C leads to a different Julia set. In other words, c / M if and only if the orbit of 0 under P c, that is to say, the sequence 0, c, c 2 + c, (c 2 + c) 2 + c,... is unbounded. It is this observation which motivates the computer graphics representing M, which one can write along the same lines as the second method for representing Julia sets presented above. It is easy to show that if c > 2, then the sequence {P n c (0)} n always diverges, so that M D 2 (0). Method for visualizing M: Fix a large integer N max (say 500). (i) Discretize the square [ 2, 2] [ 2, 2] finely enough. (ii) For each gridpoint c, compute terms of the sequence P n c (0) until they exit D 2 (0) or until n reaches N max. Define n(c) as either the smallest integer less than N max such that P n c (0) > 2, or N max if the orbit of c remains in D 2 (0) until N max. (iii) Visualize n(c).

86 86 Figure 16: The Mandelbrot set M. As c follows the plain arrow, the system undergoes a periodoubling bifurcation. As c follows the dashed arrow, the system undergoes a period-tripling bifurcation. Some facts about the Mandelbrot set. As interior points of the Mandelbrot set are such that the orbit of 0 is bounded, we may wonder what its behavior is, in particular, how many accumulation points the sequence {Pc n (0)} n has. When 0 is attracted to a k-cycle, that sequence has exactly k accumulation points. In general, we define H a hyperbolic component of M a component of M where the orbit of the point 0 is attracted to a cycle of fixed period. One can show by calculation that 0 is attracted to a fixed point (or 1-cycle) if and only if c belongs to the largest subset denoted W 0 on Fig. 16, so-called the main cardioid. We denote this set W 0, of boundary equation ρ W0 (t) = e2πit, t [0, 1]. 2 e4πit 4 ( The parameter value c = ρ 1 W0 2) = 3 4 is where a period-doubling bifurcation occurs, as we have seen in the lectures on one-dimensional dynamics (which corresponds to when c is on the realaxis), what we analyzed about the family F c (x) = x 2 + c, in particular the bifurcation diagram, tells us a great deal here!), see Figure 17. When c is slightly to the left of 3 4, we are in another hyperbolic component (so-called W 1 ) where the point 0 is attracted to a 2-cycle. The multiplier of 2 this 2-cycle has precise expression 4(c + 1) so that this multiplier has modulus less than 1 if and only if c D 1 ( 1), in particular we have W 1/2 = D 1 ( 1). At the boundary of components such as 4 4 W 0 or W 1, there are typically neutral cycles happening, since these are the bordeline cases where 2 a cycle passes from attracting to repelling and another passes from repelling to attracting. In the same way that a bifurcation ( occurs ) at c = γ W0 ( 1 2 ), one can show that a period q-doubling bifurcation occurs at each point γ p W0 q : when c changes from being inside the big cardioid to

87 87 Figure 17: Near the point c = 3 4. Top, left to right: Julia set of P c for c real, taking values c < 3 4, c = and c > 4. We see on the left that the orbit of 0 converges to a 2-cycle while on the right, it converges to a single point. Bottom: c = i, outside the Mandelbrot set. The Julia set is completely disconnected and all orbits outside J eventually diverge to. However they can take arbitrarily long to do so, this si why we must increase N max in order to visualize additional structure (left is N max = 100, right is N max = 500). In either picture, the constant red regions no longer means that the corresponding seeds have bounded orbits; rather, N max was too small to see these seeds reach the region C D 2 (0).

88 88 inside the hyperbolic component W p, the point 0 changes from being attracted to a fixed point to q being attracted to a q-cycle. Figure 18: Left-middle: near the period 4-doubling bifurcation at c = γ W0 ( 1 4). Right: Example of a siegel disk, where c = γ W0 (t) for some irrational t. At points γ W0 (t) where t is irrational, in the same way that the sequence {e iαn } n is dense in the unit circle when α is irrational, the orbit of 0 seems to have infinitely many accumulation points inside the Julia set. This is the case where the Julia set is so-called a Siegel disk, see example Figure 18, right. There are many more exciting facts about the Mandelbrot set (including things that yet remain to be proved!) and the interested reader can take a look at the article The Mandelbrot set by Bodil Branner in [DK] for more interesting details. Additional properties. Against expectations and despite the fact that the Mandelbrot set looks like an agglomerate of cardiods and disks of various sizes, it turns out that it is not (quasi-)selfsimilar, unlike Julia sets. The smaller copies each have their own set of decorations which differ from the other copies. On the other hand, sets in the shape of a Mandelbrot set appear in many other one-parameter families of complex dynamics, including higher-order polynomials, a property that gives the Mandelbrot set the attribute to be universal. Let us give an example of this last fact. For ρ C, consider the family of cubics P ρ (z) = z(z 1)(z ρ) and define the Newton map N ρ (z) = z Pρ(z) P, the map to be iterated over. We know ρ(z) that N ρ has three superattracting fixed points at 0, 1 and ρ, and since they are superattracting, this precisely means that they are also critical, and are attracted by their respective basins. However, there is a fourth critical point 13 with value z c = 1+ρ 3, and one may wonder into what basin of attraction its orbit falls. For each value of ρ, we can then assign a color to the pixel ρ depending on what basin of attracting it falls into, though we may notice that some values are in fact attracted to a fourth attracting cycle. What s more is that such regions in the ρ-plane take the form of Mandelbrot sets, for which the previous discussion adapts: the hyperbolic component of this Mandelbrot set in which ρ lies will determine the period of the additional cycle to which z c is attracted. 13 When P ρ has simple roots, the equation N ρ(z) = 0 is equivalent to P ρ(z) = 0 or P ρ (z) = 0. This second equation gives us the last critical point.

89 89 There are infinitely many such Mandelbrot sets inside the ρ-plane, and whenever ρ is in one of them, the Julia set J (N ρ ) in the z-plane contains small features reminiscent of the Julia set of some quadratic polynomial! Figure 19: Left to right: zooms into the ρ-plane of the family of Newton maps of P ρ (z) = z(z 1)(z ρ). Pixels in yellow/turquoise/light blue corresponds to cases where z c = 1+ρ 3 is attracted to the fixed point ρ/1/0, respectively. The dark blue pixels are when there is a fourth attracting cycle, to which z c is attracted. Figure 20: Right: ρ-plane with a parameter value ρ inside the hyperbolic component of an attracting 3-cycle. Left: the corresponding z-plane, partitioned into the three basins of attraction of the roots of P ρ, as well as the additional basin of attraction of a 3-cycle, whose boundary resembles the Julia set of some quadratic polynomial z z 2 + c (compare to Fig. 14, bottom). Zooming into Mandelbrot. Although it is not self-similar, the Mandelbrot set is still very complicated at every scale, and one may want to zoom into it at arbitrarily fine scales. Using a finer and finer grid requires increasing the precision of the computations, which in turn become computationally more intensive. See [Zoom] for an example of deep zoom into the Mandelbrot set, of a factor , taking months to render. For the record the biggest number in physical ranges is in the order of (size of a proton to the universe).

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