Complex Analysis. Xue-Mei Li

Transcription

1 Complex Analysis Xue-Mei Li May 6, 206

2 Contents Complex Differentiation 6. The Complex Plane Complex Functions of One Variables Complex Linear Functions Complex Differentiation The and operator Harmonic Functions Holomorphic Functions Rules of Differentiation The Riemann Sphere and Möbius Transforms 6 2. Conformal Mappings Möbius Transforms (Lecture 5) The Extended Complex Plane Properties of Möbius Transforms The Riemann Sphere and Stereographic Projection (lecture 7) Power Series Power series is holomorphic in its disc of convergence Analytic Continuation (Lecture 8) The Exponential and Trigonometric Functions The Logarithmic Function and Power Function (Lecture 9) Complex Integration 3 4. Complex Integration Integration Along a Curve (lecture 9) Existence of Primitives (Lecture 9) Goursat s Lemma Cauchy s Theorem: simply connected domains (Lecture 2) Supplementary Cauchy s Integral Formula Keyhole Operation and Other Techniques (Lecture 2) Cauchy s Integral Formula Taylor Expansion, Cauchy s Derivative Formulas Estimates, Liouville s Thm and Morera s Thm Supplementary Locally Uniform Convergent Sequence of Holomorphic Functions.. 48

3 5.6 Schwartz Reflection Principle (Lecture 5) The Fundamental Theorem of Algebra Zero s of Analytic Functions Uniqueness of Analytic Continuation (Lecture 7) Supplementary Laurent Series and Singularities Laurent Series Development (Lecture 7-8) Classification of Isolated Singularities (Lecture 9) Poles Essential Singularities Meromorphic Function Supplementary Winding Numbers and the Residue Theorem The Index of a Closed Curve (Lecture 22) Supplementary The Residue Theorem (Lecture 23) Compute Real Integrals Fundamental Theorems The Argument Principle (Lecture 24) Rouché s Theorem Supplementary The Open mapping Theorem (Lecture 25) Supplementary Bi-holomorphic Maps on the Disc (lecture 25-26) The Riemann Mapping Theorem 8 9. Hurwitz s Theorem (lecture 26) Family of Holomorphic Functions (Lecture 28) The Riemann Mapping Theorem (Lecture 28-29) Supplementary Special Functions Constructing Holomorphic Functions by Integration (Lecture 30) The Gamma function The zeta function(lecture 30)

4 Prologue This is the lecture notes for the third year undergraduate module: MA3B8. If you need not be motivated, skip this section. Complex Analysis is concerned with the study of complex number valued functions with complex number as domain. Let f : C C be such a function. What can we say about it? Where do we use such an analysis? The complex number i = appears in Fourier Transform, an important tool in analysis and engineering, and in the Schrödinger equation, i ψ t = 2 2 ψ 2m 2 + V (x)ψ(t, x), x a fundamental equation of physics, that describes how a wave function of a physical system evolves. Complex Differentiation is a very important concept, this is allured to by the fact that a number of terminologies are associated with complex differentiable. A function, complex differentiable on its domain, has two other names: a holomorphic map and an analytic function, reflecting the original approach. The first meant the function is complex differentiable at every point, and the latter refers to functions with a power series expansion at every point. The beauty is that the two concepts are equivalent. A complex valued function defined on the whole complex domain is an entire function. Quotients of entire functions are Meromorphic functions on the whole plane. A map is conformal at a point if it preserves the angle between two tangent vectors at that point. A complex differentiable function is conformal at any point where its derivative does not vanish. Bi-holomorphic functions, a bi-jective holomorphic function between two regions, are conformal in the sense they preserve angles. Often by conformal maps people mean bi-holomorphic maps. Conformal maps are the building blocks in Conformal Field Theory. It is conjectured that 2D statistical models at criticality are conformal invariant. An exciting development is SLE, evolved from the Loewner differential equation describing evolutions of conformal maps. The Schramann-Loewner Evolution (also known as Stochastic Loewner Evolution, abbreviated as SLE ) has been identified to describe the limits of a number of lattice models in statistical mechanics. Two mathematicians, W. Werner and S. Smirnov, have been awarded the Fields medals for their works on and related to SLE. Complex valued functions are built into the definition for Fourier transforms. For f : R R, ˆf(k) = e ikx dx, k R. 2π Fourier transform extends the concept of Fourier series for period functions, is an important tool in analysis and in image and sound processing, and is widely used in electrical engineering. 3

5 A well known function in number theory is the Riemann zeta-function, ζ(s) = n= n s. The interests in the Riemann-zeta function began with Euler who discovered that the Riemann zeta function can be related to the study of prime numbers. ζ(s) = Π p s. The product on the right hand side is over all prime numbers: Π p s = 2 s 3 s 5 s 7 s s... p s.... The Riemann-zeta function is clearly well defined for s > and extends to all complex numbers except s =, a procedure known as the analytic /meromorphic continuation of a real analytic function. Riemann was interested in the following question: how many prime number are below a given number x? Denote this number π(x). Riemann found an explicit formula for π(x) in his 859 paper in terms of a sum over the zeros of ζ. The Riemann hypothesis states that all non-trivial zeros of the Riemann zeta function lie on the critical line s = 2. The Clay institute in Canada has offered a prize of million dollars for solving this problem. In symplectic geometry, symplectic manifolds are often studied together with a complex structure. The space C is a role model for symplectic manifold. A 2-dimensional symplectic manifold is a space that looks locally like a piece of R 2 and has a symplectic form, which we do not define here. We may impose in addition a complex structure J x at each point of x M. The complex structure J x is essentially a matrix s.t. Jx 2 is the identity and defines a complex structure and leads to the concept of Khäler manifolds. Finally we should mention that complex analysis is an important tool in combinatorial enumeration problems: analysis of analytic or meromorphic generating functions provides means for estimating the coefficients of its series expansions and estimates for the size of discrete structures. Topics Holomorphic Functions, meromorphic functions, poles, zeros, winding numbers (rotation number/index) of a closed curve, closed curves homologous to zero, closed curves homotopic to zero, classification of isolated singularities, analytical continuation, Conformal mappings, Riemann spheres, special functions and maps. Main Theorems: Goursat s theorem, Cauchy s theorem, Cauchy s derivative formulas, Cauchy s integral formula for curves homologous to zero, Weirerstrass Theorem, The Argument principle, Rouché s theorem, Open Mapping Theorem, Maximum modulus principle, Schwartz s lemma, Mantel s Theorem, Hürwitz s theorem, and the Riemann Mapping Theorem. References L. V. Ahlfors, Complex Analysis, Third Edition, Mc Graw-Hill, Inc. (979) 4

6 J. B. Conway. Functions of one complex variables. T. Gamelin. Complex Analysis, Springer. (200) E. Hairer, G. Wanner, Analyse Complexe et Séries de Fourier. G. J. O. Jameson. A First course on complex functions. Chapman and Hall, (970). E. M. Stein and R. Shakarchi. Complex Analysis. Princeton University Press. (2003) Acknowledgement. I would like to thank E. Hairer and G. Wanner for the figures in this note. 5

7 Chapter Complex Differentiation. The Complex Plane The complex plane C = {x + iy : x, y R} is a field with addition and multiplication, on which is also defined the complex conjugation x + iy = x iy and modulus (also called absolute value) z = z z = x 2 + y 2. It is a vector space over R and over C with the norm z z 2. We will frequently treat C as a metric space, with distance d(z, z 2 ) = z z 2, and so we understand that a sequence of complex numbers z n converges to a complex number z is meant by that the distance z n z converges to zero. The space C with the above mentioned distance is a complete metric space and so a sequence converges if and only if it is a Cauchy sequence. Since z n z 2 = Re(z n ) Re(z) 2 + Im(z n ) Im(z) 2, z n converges to z if and only if the real parts of (z n ) converge to the real part of z and the imaginary parts of (z n ) converge to the imaginary part of z. In polar Coordinates z C can be written as z = re iθ where r = z and θ is a real number, called the argument. We note specially Euler s formula: e iθ = cos(θ) + i sin(θ), so arg z is a multi valued function. It is standard to take the principal value π < Argz π, a rather arbitrary choice. Since e 2πik = for k an integer, the nth root function is multi-valued. If the nth roots of the unity, then ω k = e 2πk n i, k = 0,,..., n, (re iθ ) n = r n e i θ n ωk..2 Complex Functions of One Variables To discuss complex differentiation of a function, we request that it is defined on a subset of the complex plane C which is open. By a set we would usually mean a subset of the 6

8 complex plane C. A set U is open if about every point in U there is a disc contained entirely in U. We further assume that the set is connected, otherwise we could treat it as a separate function on each connected subset. A subset of C is connected if any two points from the subset can be connected by a continuous curve which lies entirely within the subset. An open set is connected if and only if it is not disconnected in the sense that it is not the union of two disjoint open sets. Definition.2. By a region we mean a connected open subset of C. By a proper region we mean an open connected subset of C that is not the whole complex plane. From now on, by a function we mean a function f : U C where U is a region. By an open disc we mean {z : z z 0 < r} where z 0 C and r > 0. A closed disc is {z : z z 0 r} where z 0 C and r 0. The unit disc centred at 0 is denoted by D = {z : z < }. Other frequently seen open sets are the deleted discs and the annulus and polygons. {z : 0 < z z 0 < r}, {z : r < z z 0 < r 2 }, Example.2. Given z 0 C, the function f : C C given by the formula f(z) = z + z 0 is said to be a translation. As a set we may wish to identify a complex number s + it with the pair of real numbers (s, t), so C is identified with R 2. Since, for z = x + iy and c = s + it, c(x + iy) = (sx ty) + i(tx + sy), the map z cz is represented by a a linear map: ( ) ( x s t y t s ) ( x y ). Multiplication by i is the same as multiply by J on the left, where ( ) 0 J =. 0 Example.2.2 Given c C, the function f(z) = cz is of the form below. For z = x + iy, c = c e iθ, ( ) ( ) ( ) x cos(θ) sin(θ) x c. y sin(θ) cos(θ) y This is the composition of a rotation by an angle θ and a scaling by c. This map preserves the angle between two vectors, i.e. it is a conformal map. 7

9 c z c w = c z Figure.: Graph by E. Hairer and G. Wanner Example.2.3 Define f(z) = z 2 whose maximal domain of definition is C. Write f = u + iv. Then u(x, y) = x 2 y 2, v(x, y) = 2xy. It is easy to see that f takes the horizontal lines y = b where b 0 to parabolas on the w plane facing right. Solve the equations: x 2 b 2 = u and v = 2xb to see u = 4b 2 v2 b 2. Also, f takes the vertical lines x = a where a 0 to parabolas on the w plane facing left. u = 4a 2 4a 2 v2. These two sets of parabolas intersect at right angles, see Figure.2. If b = 0, the line y = 0 is mapped to the right half of the real axis; the line x = 0 is mapped to the left half of the real axis. We observe that at 0, these two curves, images of the real and complex line from the domain space, fail to intersects with each other at a right angle. (0 is the only point at which f = 0, explaining the orthogonality and failing of the orthogonality where the image curves meet, to which we return later.) The function f(z) = z 2 is not injective. It takes the line y = b and y = b to the same image. To see this better let us use polar coordinates. Then f(re iθ ) = r 2 e 2iθ. When restricted to the positive half complex plane, f is injective. In fact, f : C + C \ [0, ) is a bijection with inverse f (w) = w. When restricted to the negative half plane f : C C \ [0, ), f is also injective with inverse f (w) = ω 2 w = w. 8

10 z = w w = z 2 0 Figure.2: Graph by E. Hairer and G. Wanner Example.2.4 Define f on C \ (, 0] by f(w) = w, the principal brach of the square root function. So f(re iθ ) = re iθ/2, π < θ < π. It has another formula: f(w) = w e i(argw/2), w C \ (, 0]. It maps the slit w plane into the right half of the z-plane. The other branch of the square root is w. It is possible to glue the two slit domains together to form a complex manifold, known as a Riemann surface, so in one sheet (chart) the function takes the value of one brach and in the other we use the other brach in a way f changes continuously as w changes. Example.2.5 The map f(z) = z, the inversion map, is defined on C \ {0}. It is easy to see that f takes circles centred at the origin to circles centres at the origin. It take the locus of the solutions of z z 0 = r to that of a circle in the w-plane, see Proposition and Example sheets. Example.2.6 (Möbius Transforms) Let a, b, c, d C where ad bc. Define f(z) = az + b cz + d. If c = 0 the domain of f is C otherwise it is C \ { d c }. Exercise.2.7. Prove that for any real number r, not, the equation z z = r z z 2 determines a circle. 2. Prove that any Möbius transform is a composition of translations, scalings, and inversions. 9

11 Later we will see that Möbius Transforms can be considered as maps on the extended complex plane, the Riemann sphere..3 Complex Linear Functions We identify R 2 with C. A function T : R 2 R 2 is real linear if for all z, z 2, z R 2, T (z + z 2 ) = T (z ) + T (z 2 ), T (rz) = rt (z), r R A map T : C C is complex linear if for all z, z 2, z C, T (z + z 2 ) = T (z ) + T (z 2 ), T (kz) = kt (z), k C Proposition.3. A real linear function T : R 2 R 2 is complex linear iff T (i) = it (). We now look at the matrix representations. Every real linear map is of the form ( ( ( ) x a b x. y) c d) y If k = s + it, the complex linear map T (z) = kz is given by ( ) ( ) ( ) x s t x T =. y t s y For every real linear map T there exists a unique pair of complex numbers λ and µ such that T (z) = λz + µ z, which is complex linear if and only if µ = 0. Furthermore, λ = ((a + ic) + i ) 2 (b + id), µ = ((a + ic) i ) 2 (b + id).4 Complex Differentiation Let f = u + iv, defined in a region U. When C is identified as R 2 we may treat u and v as real valued functions on R 2. In this way f is an R 2 valued function of two real variables x and y. Then f is (real) differentiable at (x 0, y 0 ) if there exists a linear map (df) (x0,y 0) : R 2 R 2 and a function φ such that ( ) ( ) x x0 f(x, y) = f(x 0, y 0 ) + (df) (x0,y 0) + φ(x, y) x x0 y y 0 (.4.) y y 0 0

12 where φ satisfies φ(x 0, y 0 ) = 0 and lim (x,y) (x0,y 0) φ(x, y) = 0. Note that ( ) x x0 = z z y y 0. 0 The linear function is represented by the Jacobian matrix. ( ) x u J f (x 0, y 0 ) = y u (x x v y v 0, y 0 ). The partial derivatives of f are denoted by ( x u x f = x v Treated as a complex function, ), y f = x f = x u + i x v, y f = y u + i y v. ( ) y u. (.4.2) y v Definition.4. A function f : U C is complex differentiable at z 0 if there exists a complex number f (z 0 ) and a function ψ with ψ(z 0 ) = 0 and lim z z0 ψ(z) = 0, such that f(z) = f(z 0 ) + (df) z0 (z z 0 ) + ψ(z) z z 0. (.4.3) The number f (z 0 ) is the derivative of f at z 0. Equivalently, f is complex differentiable at z 0 with derivative f (z 0 ) if and only if f f(w + z 0 ) f(z 0 ) (z 0 ) = lim. w 0 w Example.4. f(z) = z is differentiable. So are any polynomials in z. This follows from and chain rules. f(z 0 + z) f(z 0 ) z = Example.4.2 The function f(z) = z is not complex differentiable. Proof Note f(z 0 + z) f(z 0 ) z = z z = {, if Im(z) = 0, if Re(z) = 0, which means lim z 0 f(z 0+z) f(z 0) z does not exist. Definition.4.2 A function is differentiable in U if it is differentiable everyewhere in U. Notation. A function f : R n R m is C r if it is r times differentiable and its partial derivatives of order less or equal to r are continuous.

13 Figure.3: Handwriting by Riemann Theorem.4.3. If f : U R is complex differentiable at z 0 = x 0 + iy 0 then f is real differentiable at (x 0, y 0 ) and the Cauchy-Riemann Equations hold at z 0 : x u = y v, y u = x v. (.4.4) Also, f (z 0 ) = x u + i x v = i ( yu + i y v). 2. If f : U C is real differentiable and satisfies the Cauchy-Riemann equation at a point (x 0, y 0 ) U then f is complex differentiable at z 0 = x 0 + iy 0. In particular, if u, v : R 2 R 2 are C functions satisfying the Cauchy-Riemann equations in U then f = u + iv is complex differentiable in U. Proof () Write f (z 0 ) = s + it. Then by the definition, (.4.3), ( ) s t f(z) = f(z 0 ) + (z z t s 0 ) + ψ(z) z z 0. This is (.4.) with ( ) s t (df) (x0,y 0) =. t s So f is real differentiable with ( ) ( ) x u y u s t =. x v y v t s Thus the Cauchy-Riemann equation follows and f (z 0 ) = s + it = x u + i x v = y v i y u. (2) We have (.4.), ( ) ( ) x x0 f(x, y) = f(x 0, y 0 ) + (df) (x0,y 0) + φ(x, y) x x0 y y 0. y y 0 By the Cauchy-Riemann equation the Jacobian matrix is the following form ( ) x u J = x v, x v x u 2

14 and represent the complex linear map: multiplication by f (z 0 ) := x u + i x v, Hence f(z) = f(z 0 ) + f (z 0 )(z z 0 ) + φ(z) z z 0. This implies f is complex differentiable at z 0. If u, v are C, then f = (u, v) is differentiable and the previous statement applies. The Cauchy Riemann equation can also be written as x f = i yf..5 The and operator Given a function f, we have This inspires the notation : f(x, y) = f ( z + z 2, z z ). 2i z = 2 ( x + i y), z = 2 ( x i y). (.5.) It is common to denote z by and z by. It is clear that z f = 0 is the Cauchy- Riemann equation We can reformulate the earlier theorem using these notations. Suppose that f is complex differentiable at z then f is real differentiable at z and, f(z) = 0, f (z) = z f(z)..6 Harmonic Functions Definition.6. A real valued function u : R 2 R is a harmonic function if u = 0 where = xx + yy is the Laplacian. Proposition.6. If u, v are C 2 functions and satisfies the Cauchy-Riemann equations x u = y v, y u = x v, then u, v are harmonic functions. Consequently u, v are C. Proof We differentiate the Cauchy-Riemann equation to see xx u = xy v, yy u = yx v. Consequently xx u + yy u = 0. Similarly, xx v + yy v = 0. From standard theory in PDE, a solution of the elliptic equation u = 0 is C. Later we see that if f is differentiable in a region, it has derivatives of all orders. So the conditions u, v C 2 can be reduced to C. 3

15 .7 Holomorphic Functions Definition.7. A function f : U C is said to be holomorphic on U if it is differentiable at every point of U; it is holomorphic at z 0 if it is holomorphic in a disc containing z 0. A function f : C C is said to be entire if it is complex differentiable at every point of C. For the next corollary we use the following, thia is also Corollary Proposition.7. A holomorphic function in a region with vanishing derivative must be a constant. Proof To see this, we first note that f has vanishing Jacobian matrix, and so it derivatives along the coordinate directions vanishes. So f is constant along any line segment parallel with x and y axis. But any two points in a region can be connected by piecewise line segments parallel to either x and y axis, and so the values of f at these two points must be the same. Example.7.2 Let U be a region in C. If f : U R is a real valued function, then f is not holomorphic in U unless f is a constant. Proof Let f = u + iv where v vanishes identically. If f is holomorphic, by the Cauchy-Riemann equation, x u = y u = 0 and f (z) = y v + i x v = 0, and f must be a constant..8 Rules of Differentiation Theorem.8. If f, g are differentiable at z 0, the derivatives indicated below exist at z 0 and the relations stated below hold when evaluated at z 0.. (kf) = kf, for any k C 2. (f + g) = f + g 3. (fg) = fg + f g 4. (f/g) = gf fg g 2 provided g(z 0 ) 0. Theorem.8.2 Suppose that g is differentiable at z 0 and f is differentiable at g(z 0 ) then the composition f g is differentiable at z 0 and (f g) (z 0 ) = f (g(z 0 )) g (z 0 ). Observe that if the Jacobian matrix of a complex differentiable function f represents complex multiplication, i.e. it is of the form ( ) x u J = y u, y u x u 4

16 then so is its inverse: J = det J Since f (z 0 ) = x u(z 0 ) i y u(z 0 ), and ( x u y u y u x u f (z 0 ) = xu(z 0 ) + i y u(z 0 ) ( x u) 2 (z 0 ) + ( y u(z 0 )) 2 = det J(z 0 ) ( xu(z 0 ) + i y u(z 0 )). ). In conclusion, J(z 0 ) represents f (z 0 ), J (z 0 ) represents f (z 0). This leads to the following theorem. Theorem.8.3 Suppose that f : U C is complex differentiable and u, v have continuous partial derivatives. Suppose f (z 0 ) 0 for some z 0 U. Then there exists a disc U around z 0 such that f : U f(u) is a bijection, f(u) is open and f : f(u) U is continuous. Furthermore f is complex differentiable on f(u) and (f ) (f(z)) = f (z), z U. Proof The first part of the statement follows from real analysis. To see f is differentiable, write w 0 = f(z 0 ), w = f(z). Since f and f are continuous, w w 0 is equivalent to z z 0. Since f (z 0 ) 0, f (w) f (w 0 ) w w 0 =. f(z) f(z 0) z z 0 Take w w 0 we see that the limit on the left hand side exists and f (w 0 ) = f (z 0 ) = f (f (w)). Remark.8.4 Later we see that if f is complex differentiable, it is infinitely differentiable. If f is one to one then f does not vanish. See section

17 Chapter 2 The Riemann Sphere and Möbius Transforms 2. Conformal Mappings Definition 2.. A parameterized curve in the complex plane is a function z : [a, b] C where [a, b] is closed interval of R. If z(t) = x(t) + iy(t) its derivative is z (t) = x (t) + iy (t). Definition 2..2 The parameterized curve : [a, b] C is smooth if z (t) exists and is continuous on [a, b]. We assume furthermore z (t) 0. The derivatives at the ends are understood to be one sided derivatives. From now on by a curve we mean a smooth curve. If z (t) does not vanish the curve has a tangent at this point, whose direction is determined by arg(z (t)). Definition 2..3 Let z : [a, b ] C and z 2 : [a 2, b 2 ] C be two smooth curves intersecting at z 0. The angle of the two curves is the angle of their derivatives at this point. They are given by the difference of the arguments of their derivatives. If z (t ) = z 2 (t 2 ) = z 0, their angle at the point z 0 is: arg(z 2(t 2 )) arg(z (t )). Definition 2..4 A map f : U C is conformal at z 0 if it preserves angles, i.e. if z and z 2 are two curves meeting at z 0, the angle from f z to f z 2 at f(z 0 ) are the same as the angle from z to z 2 at z 0. Example 2.. The linear map f(z) = kz where k 0, is a conformal map as it is composed of scaling by k and rotating by the angle arg(k). c.f.example.2.2 Example 2..2 f(z) = z is not a conformal map. This map reverses orientation. Theorem 2..3 If f : U C is holomorphic at z 0 and f (z 0 ) 0, then f is conformal at z 0. 6

18 Proof Let z : [a, b ] C and z 2 : [a 2, b 2 ] C be two smooth curves intersecting at z 0 : for t [a, b ] and t 2 [a 2, b 2 ], z (t ) = z 2 (t 2 ) = z 0. Let (t) = f z (t) and 2 (t) = f z 2 (t). Since f (z 0 ) does not vanish, and 2 have well defined tangents which are: (t ) = d dt t=t f z = f (z (t ))z (t ) = f (z 0 )z (t ) (t 2 ) = d dt t=t 2 f z 2 = f (z 2 (t 2 ))z 2(t 2 ) = f (z 0 )z 2(t 2 ). Multiply z (t ) and z (t ) by the non-zero complex number f (z 0 ) preserves angles between the two vectors, as well as their orientation, c.f. Example 2.., so the angle from to 2 at f(z 0 ) is the same as the angle from z to z 2 at z 0. Note also, arg( (t )) = arg(f (z 0 )) + arg(z (t )). 2.2 Möbius Transforms (Lecture 5) A polynomial P (z) = a 0 + a z +... a n z n, where a i C, is an entire function. The roots z n are the zero s of P. If there are exactly r roots coincide, this root is said to have order r. In light of Theorem 2..3 it is interesting to know where lie the zero s of P (z). By the fundamental theorem of Algebra, which we prove later (Theorem 5.7.2), P (z) = 0 has a complete factorisation: P (z) = a n (z z )... (z z n ). Suppose that P and Q are two polynomials without common factors and define the rational function f(z) = P (z) Q(z). Then f is defined and is complex differentiable everywhere except at the zeros of Q. The zero s of Q are the poles of f. We now look at rational functions with one pole and one zero. Definition 2.2. The following collection of maps are Möbius transforms { } az + b : ad bc 0, a, b, c, d, C. cz + d If ad bc = 0, az+b az+b cz+d is a constant function, and are hence excluded. If f(z) = cz+d is a Möbius transform, its maximal domain is: C \ { d c }. Since f is a conformal map. f (z) = ad bc (cz + d) 2 0, 7

19 2.2. The Extended Complex Plane To make the statements neat we add a point at infinity to C and define the extended complex plane to be C = C { } with the convention: 0 =, = 0, a + =, a =, and for a 0, a = a = Properties of Möbius Transforms Let f(z) = az+b cz+d. We extend the Möbius transform f from C to C by defining: f( d c ) =, f( ) = a c if c 0. If c = 0, f(z) = f (az + b) is defined on the whole plane, then we define The function f has an inverse f( ) =, if c = 0. f (w) = dw b cw + a. Note that multiply a, b, c, d, by a non-zero number λ does not change the function f(z) = az + b azλ + bλ = cz + d cλz + dλ. Hence we may eliminate one parameter and assume that ad bc =. We define { } az + b M = : ad bc =, a, b, c, d, C. cz + d Theorem 2.2. The set M of Möbius transforms is a group under composition. Each Möbius transform is a composition of the following maps: () translation: z z + a for some complex number a; (2) composition of scaling and rotation: (3) Inversion: z z. Proof For the group we check the following: z kz, some k C, k 0. f(z) = z is the identity. (a =, b = c = 0, d = ) If f(z) = az+b cz+d M, then f (w) = 8 dw b cw + a M.

20 If f(z) = az+b āz+ b Az+B cz+d M and g(z) = M. Then f g = cz+ d Cz+D M where the complex numbers A, B, C, D are given by ( ) ( ) (ā ) A B a b b =. C D c d c d For the second part of the statement, if c = 0, az+b d = a d z + b d. If c 0, f(z) = az + b cz + d = a c bc ad + c 2 z + d. c Example The map f(z) = z+ z is called the Cayley transform. It takes C \ {} to itself, f : C \ {} C \ {} is a bijection and f = f. Let us consider f as a map on C by setting f() =, f( ) =. Note f(x + iy) = x2 + y 2 (x ) 2 + y 2 + 2y (x ) 2 + y 2 i. Let = {x 2 + y 2 = } with + and denote respectively the upper and lower half of the circle. Then, f sends {, 0, } to {0,, } respectively. f sends the upper circle to the lower half of the imaginary axis. f sends x-axis to x-axis. It send the x-axis within the unit disc to the negative x-axis. f sends the upper half of the unit disc to the third quadrant. f sends the lower circle to the upper half of the imaginary axis. f sends the lower half of the unit disc to the second quadrant. f sends the exterior of the unit circle to the right half of the plane. If z 2, z 3, z 4 are distinctive points in C we associate to it the Möbius transform f(z) = z z 3 z z 4 / z 2 z 3 z 2 z 4 = (z z 3)(z 2 z 4 ) (z z 4 )(z 2 z 3 ), if z 2, z 3, z 4 C. If one of these points is the point at infinity the map is interpreted as following: Note that if z 2, z 3, z 4 C, z z 3, if z 2 = z z 4 z 2 z 4 f(z) =, if z 3 = z z 4 z z 3, if z 4 =. z 2 z 3 f(z 2 ) =, f(z 3 ) = 0, f(z 4 ) =. 9

21 Also, If z 2 =, f( ) =, f(z 3 ) = 0, f(z 4 ) =, If z 3 =, f(z 2 ) =, f( ) = 0, f(z 4 ) =, If z 4 =, f(z 2 ) =, f(z 3 ) = 0, f( ) =. We denote by F z2z 3z 4 this map. Note it sends {z 2, z 3, z 4 } to {, 0, }. Lemma A Möbius transform can have at most two fixed points unless f(z) is the identity map. Proof We solve for az+b cz+d = z, equivalently cz2 + (d a)z b = 0. This has at most two solutions(use polynomial long division/ the Euclidean algorithm). Proposition For any two sets of distinctive complex numbers {z 2, z 3, z 4 } and {w 2, w 3, w 4 } in C, there exists a unique Möbius transform taking z i to w i for i = 2, 3, 4. Proof We know F z2z 3z 4 takes {z 2, z 3, z 4 } to {, 0, }, and the inverse map of F w2w 3w 4 takes {, 0, } to {w 2, w 3, w 4 }. The composition Fw 2w 3w 4 F z2z 3z 4 takes {z 2, z 3, z 4 } to {w 2, w 3, w 4 }. To prove this map is unique, suppose f, g are two Möbiums transform sending {z 2, z 3, z 4 } to {w 2, w 3, w 4 }. Then f g (w i ) = f(z i ) = w i. The Möbius transform f g has three fixed points: w, w 2, w 3. By Lemma 2.2.3, f g is the identity map and f = g identically. Corollary For any distinctive complex numbers {z 2, z 3, z 4 } in C, the Möbius transform F z2,z 3,z 4 is the only Möbius transform that takes {z 2, z 3, z 4 } to {, 0, }. Proposition Let r, c be real numbers, k C. Then the equation represents a line if r = 0 and k 0. a circle if r 0, and k 2 rc. r z 2 + kz + k z + c = 0 The circle equation is z + k r = r k 2 rc, whose locus is an emptyset if r 0 and k 2 < rc. This is clear by expanding z = x + iy in x and y. Definition The locus of the points of r z 2 kz k z + c = 0, if non-empty, is called a circleline. We see later this definition is not merely a simplification of terminologies. Both circles and extended lines in the plane correspond to circles in the Riemann sphere. Lemma Let R be a real number, z, z 2 complex numbers. The locus of the equation z z = r z z 2 20

22 represents a circle if r. If z = x + iy, z = x + iy and z 2 = x 2 + iy 2 then If r =, the equation is (x x ) 2 + (y y ) 2 = r(x x 2 ) 2 + r(y y 2 ) 2. 2(x 2 x )x + 2(y 2 y )y = z 2 2 z 2, representing a line if z z 2. It is the set of points which are equi-distance from z and z 2, i.e. the line perpendicular to the line segment [a, b] and passing its mid-point. Proposition A Möbius transform maps a circleline to a circleline. Proof Since a Möbius transform is the composition of translation, multiplication by a non-zero complex number and inversion we only need to prove it for each of these maps. The inverse of such transformations are of the same type. A translation, z z + a, takes a circleline to a circleline: the image of r z 2 + kz + k z + C = 0, in the z-plane is precisely the locus of the equation below in the w-plane: i.e. r w a 2 + k(w a) + k( w ā) + C = 0, r w 2 + (k a)w + (k a) w + r a 2 ( ka + kā) + C = 0. Note that r a 2 ( ka + kā) + C is a real number. Multiplication by a complex number is a composition of scaling with rotation, it clearly takes a circleline to a circleline. Finally we work with the inversion z z. It takes z = r to z = r trivially. Let a 0. If w is in the image of z a = r then w a = r, i.e. w a = r a w which is a circleline, by Lemma Let us take a line kz +k z +C = 0. The equation of its image w = z satisfies k w +k w +C = 0 which is equivalent to k w +kw +C w 2 = 0 representing a circle if C 0 and a line otherwise. Exercise Given r, c R and k C, and the equation r z 2 kz k z + c = 0. Identify its image under the transform w = k z where k is a non-zero complex number. Definition The cross ratio of z, z 2, z 3, z 4, denoted by [z, z 2, z 3, z 4 ], is the complex number: [z, z 2, z 3, z 4 ] := F z2z 3z 4 (z ). In other words, it is [z, z 2, z 3, z 4 ] = (z z 3 )(z 2 z 4 ) (z z 4 )(z 2 z 3 ), interpreted appropriately if one of them is. Proposition Let z, z 2, z 3, z 4 be distinct points in C. Then [z, z 2, z 3, z 4 ] is a real number if the four points lie in a circle or on the extended line R { }. 2

23 Proof If [z, z 2, z 3, z 4 ] is a real number, F z2,z 3,z 4 maps the four points z, z 2, z 3, z 4 to respectively [z, z 2, z 3, z 4 ],, 0,, all on the extended x-axis. The map (F z2,z 3,z 4 ) takes the 4 points [z, z 2, z 3, z 4 ],, 0, back to z, z 2, z 3, z 4. Note a Möbius transform takes the extended line to a circleline, so the four points lie in a circleline. If the four points lie in a circleline, then the map F z2,z 3,z 4 takes the circleline to a circleline. This will be the line determined by (, 0, ), the x-axis. Hence F z2,z 3,z 4 (z ) must be a real number. 2.3 The Riemann Sphere and Stereographic Projection (lecture 7) The purpose of the section is to give a concrete geometric representation of the extended plane as the Riemann sphere. In particular we observe that the point at infinity is just represented as a point in the sphere. Let us denote by S 2 the unit sphere in R 3 : S 2 = {(X, Y, Z) : X 2 + Y 2 + Z 2 = }. We fix the north pole N = (0, 0, ) and associate with each P on S 2 \{N} with a point π(p ) on the plane which is the intersection of the line from N to P with the plane. Proposition 2.3. The stereographic projection from S 2 C is : The inverse map is given by π((x, Y, Z)) = X + iy Z, π(n) =. (2.3.) π (z) = ( 2Re(z) z 2 +, 2Im(z) z 2 +, z 2 z 2 ). (2.3.2) + Proof Suppose that P = (X, Y, Z), write (x, y, 0) = π(p ). The line equation connecting N, P and π(p ) is given by: Setting z = 0 we see (x, y, z) = (0, 0, ) + t(x 0, Y 0, Z ). t = Z, x = tx = X Z, y = ty = Y Z, (2.3.3) 22

24 proving π((x, Y, Z)) = X+iY Z. Let z = x + iy be a point in C, we find its inverse π (z). We use t 2 (X 2 + Y 2 + Z 2 ) = t 2. Thus Finally x 2 + y 2 = t 2 ( Z 2 ) = Z2 ( Z) 2 = + Z Z. Z = x2 + y 2 x 2 + y 2 + = z 2 z 2 +. By t = Z and (2.3.3), we see X = x( Z) = 2x The following is an easy exercise. z 2 +, Y = y( Z) = 2y z 2 +. Proposition The antipodal point to a point (X, Y, Z) in S 2 is ( X, Y, Z). If z C corresponds to a point in S 3 then z corresponds to the antipodal point in S2. Definition 2.3. If z, z 2 C we define the stereographic distance to be d(z, z 2 ) = π (z ) π (z 2 ). If p = (X, Y, Z) and p = (X, Y, Z ) are points in the sphere, their distance is: P P = X X 2 + Y Y 2 + Z Z 2 = 2 2(XX + Y Y + ZZ ) (2.3.4) If z =, then π (z ) = (0, 0, ), X = 0, Y = 0 and Z =. Consequently, d(z, ) = 2 2 z 2 z 2 + = 2 z 2 +. This agrees with the intuition, z means z. If z, z C, apply (2.3.4), and use (2.3.2) we see ( ) (d(z, z )) 2 (z + z) = 2 2 z 2 + (z + z z z z ) z z i z 2 + i z z 2 z 2 + z 2 z 2 + Since ( z 2 + )( z 2 + ) ( z 2 )( z 2 ) = 2 z z 2 (z + z)(z + z ) (z z)(z z ) = 2zz + 2 zz. Also, z 2 + z 2 zz zz = (z z )( z z ) = z z 2. Cleaning up the right hand side we obtain: d(z, z 2 z z ) = z 2 + z 2 +. Definition The space S 2 is the Riemann sphere. A circle on S 2 is the intersection of a plane with S 2. Proposition A circle on S 2 corresponds to a circle or a line on C. 23

25 Proof let us take a plane: AX + BY + CZ + D = 0 where A, B, C, D are real numbers. Note that the north pole passes through the plane if and only C + D = 0. Let z = x+iy C. Then a point π 2x (z) = ( z 2 +, 2y z 2 +, z 2 z 2 + ) on S2 satisfies the plane equation if and only if Rearrange the equation: 2xA + 2BY + ( z 2 )C + ( z 2 + )D = 0. (C + D)(x 2 + y 2 ) + 2xA + 2By + (D C) = 0, (2.3.5) which represents a circle in the plane or an empty set when C + D 0. If the plane intersects with S 2, it is not empty and so is a circle. If C + D = 0 this is a line on the plane. Let us consider a circle or an extended line in C. It is of the form: Ã(x 2 + y 2 ) + Bx + Cy + D = 0 (2.3.6) where Ã, B, C, D are real numbers. Let us solve for A, B, C, D: C + D = Ã, 2A = B, 2B = C, D C = D. Then (2.3.6) is equivalent to (2.3.5) which means the corresponding points of the circleline on the plane satisfies AX + BY + CZ + D = 0, and their image by π line on a circle in S 2. If the plane passes through the origin we have a great circle. This is so if and only if D = 0 and we have (x 2 + y 2 ) + 2A C x + 2B C y =. The plane passes through the north pole if and only if C + D = 0 in which case the circle projects to a line. 24

26 Chapter 3 Power Series Definition 3.0. A series of complex numbers n=0 a n is said to converge if the partial sum N n=0 a n converge. It is said to converge absolutely if n=0 a n converges. Evidently n=0 a n is convergent is equivalent to both n=0 Re(a n) and n=0 Im(a n) converge. Proposition 3.0. If n=0 a n converges absolutely, then it is convergent. Just note that Re(a n ) a n and Im(a n ) a n. Follow this with the standard comparison test. 3. Power series is holomorphic in its disc of convergence Let us consider a power series n=0 a n(z z 0 ) n where a n, z 0 and z are complex numbers. For simplicity let us take z 0 = 0. Theorem 3.. Let n=0 a nz n be a power series where a n C. There exists R [0, ], such that the following holds: () If z < R, the series converges absolutely. (2) If z > R, the series diverges. Moreover, there is Hadamard s formula: R = lim sup( a n ) n n with the convention = 0 and 0 =. The region { z < R} is called the disc of convergence and R its radius of convergence. Proof Suppose A = lim sup n ( a n ) n is such that 0 < A <. Then there exists N such that for n N, a n n A. If z < A, then there exists δ > 0 such that z < A+δ and for n N, a n n z A A+δ <, and n=0 a n z n is convergent. If 25

27 z > A, there exists 0 < δ < A such that z > A δ and a n n z A A δ > for n N, it follows that n=0 a nz n is divergent. If lim sup n ( a n ) n =, then for any non-zero z, a n z n does not converge to 0 as n and the power series is divergent for all z 0. ( There is a sequence a nk with a nk n k > 2 z ). If lim sup n ( a n ) n = 0, then for any z and for any 0 < ɛ < 2 z there exists N such that a n n ɛ, for all n N, and a n z n ɛ n z n 2 n. The power series converges absolutely for any z. By composing with translation z z z 0, we may translate the statement of the theorem from the power series n=0 a n z n to the power series n=0 a n z z 0 n. Theorem 3..2 The power series f(z) = n=0 a n(z z 0 ) n defines a holomorphic function in its disc of convergence. Furthermore, f (z) = n a n (z z 0 ) n n= and f has the same radius of convergence as f. Proof If R is the radius of convergence for f, then using Hadamard s formula we see the radius of convergence for n= n a n(z z 0 ) n is R. Take z from its disc of convergence, {z : z z 0 < R}. Define f N (z) = N a n (z z 0 ) n, f N(z) = n=0 N n= n a n (z z 0 ) n. Then for h C, f(z + h) f(z) n a n (z z 0 ) n h n= f N (z + h) f N (z) f N (z) h + a n (z + h z 0 ) n a n (z z 0 ) n h + n a n (z z 0 ) n. n=n n=n+ The last term on the right hand side is the remainder term of the convergent series n= n a n(z z 0 ) n. Given ɛ > 0, there exists N 0 such that if n N 0, this last term is less than ɛ/3. Furthermore there exists a number δ 0 > 0 and 0 < A < R such that z + h z 0 < A for h δ 0. In the following we use the identity: a n b n = (a b)(a n + a n 2 b b n ), a n (z + h z 0 ) n a n (z z 0 ) n h a n (z + h z 0 ) n (z z 0 ) n h n=n+ 26 n=n+

28 n=n+ n=n+ a n (z + h z0 ) n + (z + h z 0 ) n 2 (z z 0 ) (z z 0 ) n a n na n. The right hand side is again the tail of a convergent series, hence there exists N > N 0 such that n=n a n na n ɛ/3. Finally, f N is a differentiable function, lim f N (z + h) f N (z) h 0 S N (z) h = 0. By choosing h sufficiently small, f N (z+h) f N (z) h S N (z) < 3ɛ. The proof is complete. Corollary 3..3 A power series function f(z) = n=0 a n(z z 0 ) n is infinitely differentiable in its disc of convergence. Furthermore a n = f (n) (z 0) n!. 3.2 Analytic Continuation (Lecture 8) Definition 3.2. A function f : U C is said to be analytic, or has a power series expansion, at z 0 U, if there exists a power series with positive radius of convergence such that f(z) = a n (z z 0 ) n, for z in a neighbourhood of z 0. n=0 We say f is analytic on U if it has a power series expansion at every point of U. Example 3.2. The function f(z) = z is analytic on C \ {0}. Note z = for all z <. It is clear f has the power series expansion at z = : ( z) n, z <. n=0 If z 0 is any non-zero number, take z with z 0 z < z 0, then z = z 0 (z 0 z) = z 0 z0 z = z 0 ( ) n (z 0 ) n (z z 0 ) n. n=0 n=0 zn The power series converges for any z with z 0 z < z 0 (in particular z 0). Hence f is analytic. Theorem The power series function f(z) = n=0 a n(z z 0 ) n is analytic in its disc of convergence D = {z : z z 0 < R}. In fact for w D, f(z) = n=0 f (n) (w) (z w) n, z D(w, R w z 0 ). n! 27

29 Proof Take w D. Note that w z 0 < R and let z satisfy z w < R w z 0. We expand the power series ( n ( ) ) f(z) = a n (z w + w z 0 ) n n = a n (z w) k (w z k 0 ) n k n=0 n=0 k=0 ( ( ) ) n = a k n (w z 0 ) n k (z w) k. k=0 n=k To justify the exchange of the order in the above computation, we bound the the partial sum and use the rearrangement of double series lemma below: N n n=0 k=0 ( n a n z w k) k w z 0 n k = n=0 N a n ( z w + w z 0 ) n n=0 a n ( z w + w z 0 ) n < a n r N <, where r is a number smaller than R. Hence the partial sum up to N is bounded by n=0 a n r N. Then f(z) = b k (z w) k, where b k = k=0 n=k n=0 ( ) n a n (w z k 0 ) n k. It is clear f (k) (w) = b k k!. Lemma (Double Series Lemma) Suppose there exists a number M such that N i=0 j=0 N a ij M Then all linear arrangements of the double series converge absolutely to the same number. a 00 + a 0 + a 02 + a = s 0 + a a + + a a = + s + a a a a = + s 2 + a a a a = + s 3 + : + : + : + : + : = = = = = v 0 + v + v 2 + v =??? (3.2.) Figure 3.: Figure From E. Hairer and G. Wanner 28

30 Definition If f : V C is a function where V is a subset of C and g : U C is an analytic function in a region U with V U. If f, g agree on V we say g is an analytic continuation of f into the region U. The set V is not required to be a region. We may wonder which functions has an analytic continuation. Example If f(x) = a n (x x 0 ) n is a real power series function with radius of convergence R. We define g(z) = a n (z x 0 ) n. Then g has the same radius of convergence R ( use Hadamard s formula for R). So g is an analytic continuation (also known as analytic extension) of f from ( R, R) to the disc {z : z < R}. Example If P (x) is a polynomial with one real variable then P (z) is its analytic continuation into C. Example Let h(z) = n=0 ( z)n, z <. Then f(z) = z is an analytic continuation of h into the punctured complex plane C \ {0}. Later we will study zeros of analytic function and conclude a function defined on any connected set containing an accumulation point can have only one analytic continuation. 3.3 The Exponential and Trigonometric Functions The exponential functions and trigonometric functions are analytic continuations of their corresponding functions on the real line. Definition 3.3. We define the following function by power series: e z = z n n!, sin(z) = ( ) n z2n+ (2n + )!, cos(z) = n=0 n=0 They are entire functions. By adding two series together we see that and Euler s formula: sin(z) = eiz e iz 2i, cos(z) = eiz + e iz, 2 n=0 ( ) n z2n (2n)!. e iz = cos(z) + i sin(z). n=0 We also define the following functions: sinh(z) = n=0 z2n+ (2n+)! and cosh(z) = z2n (2n)!. Note that sinh(z) = ez e z 2 and cosh(z) = ez +e z 2. Most properties for the corresponding real trigonometric functions are inherited by the complex valued trigonometric functions. For example the zeros of sin(z) are at nπ. But sin(z) is not a bounded function, nor is cos(z). 29

31 Theorem 3.3. A power series f(z) = n=0 zn n! satisfies for all z, w, z + w. In particular, f(z + w) = f(z)f(w) e 2kπi =, e z+2kπi = e z, k = 0, ±, ±2,.... Note that e x+iy = e x e iy and e iy for y ( π, π) traces out a circle without the point on the left real axis. Let U = {z : π < Imz < π} Then is a bijection. e z : U C \ {re ±iπ : r 0} Figure 3.2: Graph by E. Hairer and G. Wanner 3.4 The Logarithmic Function and Power Function (Lecture 9) Definition 3.4. The principal branch of the logarithm is the inverse of e z on the slit plane C \ {re ±iπ : r 0}. log(z) = log z + i arg(z), arg(z) ( π, π). Theorem 3.4. The logarithmic function defined above is holomorphic on its domain of definition and (log z) = z. Proof Apply Theorem.8.2 to the exponential function from U to the slit domain. Other branches of log z include: log z = log z +i arg(z)+2kπ, arg(z) ( π, π). Definition For λ C we define z λ = e λ log z. 30

32 Chapter 4 Complex Integration If a continuous function has a primitive in a region, then its integral along any closed piecewise smooth curve vanishes. The converse holds in a star region (more generally, in a simply connected region): if a continuous function integrate to zero along any triangle inside the region, then it has a primitive. Goursat s theorem states that the integral of a holomorphic function in a region indeed integrate to zero along any triangle who and whose interior is contained in the region. From this we see Cauchy s theorem for a star region: if f is holomorphic in a star region, then it integrates to zero along any closed smooth curve whose interior is contained entirely in the region. Cauchy s theorem is in fact valid for any simply connected region. Every point in a region U has a disc around it, contained entirely in U. So every point in U has a star region neighbourhood (i.e. a region is locally simply connected ). In another word, a holomorphic function integrates to zero along any closed smooth curve with sufficiently small enclosure. The distinction between the local and the global null integral property relates to homotopy theory as well as to de Rham s cohomology theory built on closed and exact differential forms, both are related to the concept of simply connectedness of a region. On the complex plane, the simply connectedness can be explained visually which is also responsible for the beauty of the theory on the plane. It is perhaps confusing in the beginning when confronted with different version s and various forms of Cauchy s formulas, the rule of thumb is the following. That a function is differentiable at a point is a local property, we could pick our disc as small as we like. A value for an integral along a closed interval is a global property: it depends on the region enclosed by the curve. 4. Complex Integration By a C function we mean a differentiable function with continuous derivative. The derivatives at the ends of an interval are one sided derivatives. Definition 4.. A parameterized smooth curve is a C function, z : [a, b] C. It is piecewise smooth if z is continuous on [a, b] and there exist t i s.t. a = t 0 < t < < t n = b s.t. z is smooth on each sub-interval [t i, t i+ ]. A piecewise smooth curve consisting of a finite number of smooth pieces, joined at the ends. We sometimes abbreviate a piecewise smooth curve to a smooth curve. A 3

33 parameterized curve has an orientation: it is the direction a point on the curve travels as the parameter t increases. Definition 4..2 Two parameterizations : [a, b] C and : [a, b ] C are equivalent if there exists a C bijection α : [a, b] [a, b ] such that α (t) > 0 and = α. The condition α (t) > 0 means orientation is preserved. The family of all equivalent parameterizations determine a smooth oriented curve. Example 4.. The circle { z z 0 = r} has the obvious parameterizations: z = z 0 + e iθ, 0 θ 2π. The orientation of the curve is anticlockwise (it is a positively oriented curve). The curve z = z 0 + e iθ, 0 θ 2π is negatively oriented. If f : [a, b] C is a continuous function and f(t) = u(t) + iv(t) then b a f(t)dt = b a u(t)dt + i b a v(t)dt. Lemma 4..2 b a f(t)dt b a f(t) dt. Proof Let θ be the principle argument of the complex number b f(t)dt. Then a b b ( b ) b f(t)dt = e iθ f(t)dt = e iθ f(t)dt = Re e iθ f(t)dt a = b a a a Re ( e iθ f(t) ) dt b a f(t) dt. a 4.2 Integration Along a Curve (lecture 9) Definition 4.2. Let f : U C be a continuous function. Let be a smooth curve contained in U with parameterization z : [a, b] C. We define the integral of f along to be: b f(z)dz = f(z(t))ż(t)dt. a 32

34 Let us write f = u + iv and z(t) = x(t) + iy(t). Then f(z(t))ż(t) = u(z(t))ẋ(t) v(z(t))ẏ(t) + i (u(z(t))ẏ(t) + v(z(t))ẋ(t)). Hence f(z)dz = b a b i ( ) u(x(t), y(t))ẋ(t) v(x(t), y(t))ẏ(t) dt + a ( ) u(x(t), y(t))ẏ(t) + v(x(t), y(t))ẋ(t) dt. The integral f(z)dz can also be defined directly by the Riemann sums: n f(z(s i ))(z(s i+ ) z(s i )) i=0 where a = s 0 < s < < s n = b is a partition on [a, b]. If the Riemann sum converges as the size of the partition converges to zero, the limit is b a f(z)dz. Proposition 4.2. The integral f(z)dz is independent of the parameterization. Proof Let z : [a, b] C and z : [a, b ] C be two parameterizations of. Let α : [a, b ] [a, b] be a C bijection such that α (t) > 0 and z = z α. Then b b a b a f( z(t)) d dt z(t)dt = = f(z α(t))ż(α(t)) α(t)dt = a f(z α(t)) d dt z(α(t))dt b a f(z(s))ż(s)ds. Definition The length of the curve is: length() = b a z (t) dt. The length of the curve is also independent of the parameterization, be an argument similar to that in the proposition above. Definition On a piecewise smooth curve, consisting of a finite number of smooth curves i, we define f(z)dz = f(z)dz. i i Theorem The following properties hold. () (k f + k 2 g)(z)dz = k f(z)dz + k 2 g(z)dz. 33

35 (2) f(z)dz = f(z)dz, where is the curve with reversed orientation, e.g. z (t) = z(a + b t). (3) f(z)dz sup f(z) length of (). z A curve is simple if it does not intersect with itself except at the end points. We would be interested in the contour of a region, e.g. the contour of a disc is the circle, traveled anticlockwise once. By a circle, we usually mean traveling along it, anti clockwise, once. Example Let be the circle z z 0 = r with positive orientation. Let n be an integer. Then { (z z 0 ) n 0, n dz = 2πi, n =. Proof Let us take the parameterization z = z 0 + re it, 0 t 2π. (z z 0 ) n dz = 2π 0 (re it ) n d dt (reit )dt 2π = r n+ i e i(n+)t dt 0 { r = n+ i 2π cos((n + )t)dt r n+ 2π sin((n + )t)dt = 0, n 0 0 i 2π dt = 2πi, n = Existence of Primitives (Lecture 9) Definition 4.3. A curve : [a, b] C is closed if (a) = (b). Definition A function f : U C where U is a region is said to have a primitive F if F : U C is holomorphic on U and F = f. The Fundamental theorem of Calculus states that if g : [a, b] R is a continuous functions, then G (x) = g(x) where G(x) = x g(t)dt. Unlike for real differentiability, the existence of a primitive is a much stronger property. To begin with, given two a points on a plane, there are many paths leading from one to the other. This is related to a path independent property. Later we see that if f has a primitive in a region, it is itself complex differentiable in this region. Rule of thumb. If you have a real valued function g of one variable x with explicit formulation which does not involve the number i, denote f the function obtained by replacing x with z. Then f is a candidate analytic continuation of g. Compute the formal anti-derivative of g and denote it by G. Now in G replace x by z and denote 34

36 the function by F. Then F is a candidate primitive for f. Once you have a primitive, check it out: is it actually complex differentiable in the desired region? If so, compute its derivative by the rule of calculus. Theorem 4.3. If f : U C is continuous and has a primitive F, if is a piecewise C curve in U begins at w and ends at w 2 then f(z)dz = F (w 2 ) F (w ). In particular, if is a closed curve then f(z)dz = 0. Proof Let z : [a, b] C be a parameterization of. We first assume that it is C and discuss the piecewise C curve later. Then b b f(z)dz = f(z(t))z (t)dt = F (z(t))z (t)dt = a b a a d dt F (z(t))dt = F (z(b)) F (z(a)) = F (w 2) F (w ). If is a piecewise smooth curve, joined by smooth curves on each subinterval of t 0 = a < t < < t n = b, then we have a telescopic sum: f(z)dz =[F (z(t n )) F (z(t n ))] + + [F (z(t 2 )) F (z(t ))] + [F (z(t )) F (z(t 0 ))] =F (z(b)) F (z(a)) = F (w 2 ) F (w ). Corollary If f : U C is holomorphic, where U is a connected open set, and f = 0 identically on U then f is a constant. Proof Let us fix a point z 0 U. Let z U be any other point. Recall U is path connected: we can connect z 0 to z by a piecewise smooth curve. Since f = 0 is a continuous function, f(z) f(z 0 ) = f (z)dz = 0. The converse to the vanishing of integral theorem is at the heart of complex integration, for which we must restrict the region. We work with a sub class of regions of the simply connected region, called the star regions. Definition A region U is said to be a star region if it has a centre C, by which we mean for all z U, the line segment belongs to U. {( t)c + tz : 0 t } 35

37 Note that z(t) = ( t)c + tz is a prameterization of the line segment from C to z. Star regions include discs, triangles, rectangles and more generally convex sets. A star region is simply connected, by which we mean any simple closed curve in that region can be continuously deformed to a point. Many regions such as polygons can be divided into star regions, which allow us to conclude statements for star regions for more general regions. Figure 4.: Graph by E. Hairer and G. Wanner Let us denote by [C, z] the line segment from C the z. Theorem (Integrability Criterion) Let U be a star region with a centre C and f : U C a continuous function. Suppose that f(ζ)dζ = 0 T for any triangle T contained entirely in U, and with C one of its vertex. Then f has a primitive in U. In particular F (z) = f(ζ)dζ, z U. [c,z] Proof The line segment [c, z] is contained in U, so F (z) = f(ζ)dζ makes [c,z] sense. Let z 0 U. Take z close to z 0 so that the triangle region with vertex z 0, z and C are contained entirely in U. Denote T this triangle. By the assumption, 36