MATH 6322, COMPLEX ANALYSIS

Transcription

1 Complex numbers: MATH 6322, COMPLEX ANALYSIS Motivating problem: you can write down equations which don t have solutions, like x 2 + = 0. Introduce a (formal) solution i, where i 2 =. Define the set C := {a + ib a, b R}. Can put a ring structure on this. (a + ib ) + (a 2 + ib 2 ) := (a + a 2 ) + i(b + b 2 ), (a + ib ) (a 2 + ib 2 ) := (a a 2 b b 2 )) + i(a b 2 + a 2 b ). One verifies that it s associative, distributive, and has the expected units. It is a field, and (a + ib) a = a 2 + b i b 2 a 2 + b. 2 Conjugate: z = a ib if z = a + ib. Norm: z := a 2 + b 2. e a+ib := e a (cos b + i sin b). We can write: z = z e iθ (this can also be seen that C = R 2 so that we can use the polar coordinate to represent (a, b). The express of z = z e iθ is called the polar form of z. Complex differential operators: Write z = ( 2 x i ), y z = ( 2 x + i ). y Let f = u + iv be a complex valued function on U which is C. If u and v are differentiable on U, then, for fixed z 0 U, f(z) f(z 0 ) = f z (z z 0) + f z ( z z 0) + o((z z 0 )). Holomorphic Functions: A complex valued function f on U is said to be holomorphic on U if f is C (U) and f = 0 on U, or equivalently, it is C (U) and it satisfies the Cauchy- z Riemann equations on U when we write f = u + iv on U. u x = v y, u y = v x ( )

2 f is said to be holomorphic at (or around) z 0 if f is holomorphic in a neighborhood of z 0. The concept of holomorphic function is also related to the complex differentiability of f: A function f defined around z 0 is said to be complex differentiable at z 0 if the complex derivative f(z 0 + h) f(z 0 ) lim h a h exists. Form (*) we get the following: Assume that f is C (U) for U C. f Then f is holomorphic (i.e. = 0 on U) if and only if f (a) exists for every z a U. Furthermore, f (z) = f z = f x = i f y on U. Warning: f (z 0 ) exists alone does not imply that f is holomorphic at z 0. For example, function f(z) = z 2 is differentiable at 0 and nowhere else, in particular, f is not holomorphic at 0. Important: (i) Goursat actually proved that f is holomorphic if f has complex derivative at each point in U (Without the assumption that f is C ). The proof will be given later. If f is holomorphic around P, then f is conformal at P (i.e. it preserves the angles and stretches in all directions (see Theorem on Page 40). Examples of holomorphic functions on C: e z := e x e iy = e x (cos y + i sin y), The (complex) sine and cosine functions: sin z = 2i (eiz e iz ), cos z = 2 (eiz + e iz ). Warning: Unlike in the real case, the complex sine and cosine are not bounded consider, for example, sin(iy) as y. u is said to be harmonic on U if u is C 2 and u = 0 on U. If f is holomporphic on U and write f = u + iv, then u and v are harmonic on U. 2

3 Solving P.D.E. on rectangular or an open disc: Lemma (Solving P.D.E. on rectangle or an open disc): Let R be a rectangle or an open disc, and if u, u 2 are C (R) with u = u 2 on R, then there is a y x function h C 2 (R) such that h x = u, h y = u 2. Moreover, if u, u 2 are real-valued, then h can also be taken as real-valued. Indeed, it is easy to check that h(x, y) = x a f(t, b)dt + y b g(x, s)ds serves our purpose. Improved version: Lemma (Solving P.D.E. on rectangle or an open disc): Let U be a rectangle or an open disc, and if u, u 2 are continuous and continuously differentiable on U\{p} with u = u 2 on U\{p}, then there is a function y x h C 2 (R) such that h x = u h, y = u 2. Moreover, if u, u 2 are real-valued, then h can also be taken as real-valued. The improved results comes from the following result from real-analysis: If F, H are continuous on (a, b) and H (x) = F (x) on (α, β)\{p} and H (p) exists. Then H (x) = F (x) on (α, β). Applications (Using the solution to the P.D.E. above): holomorphic v.s. Harmonic, Anti-derivatives (Primitives) for holomorphic: holomorphic v.s. Harmonic: If f is holomorphic on U and write f = u + iv, then u and v are harmonic on U. On the other hand, if u is a real-valued harmonic function on U where U is a rectangle or an open disc, then there is a holomorphic f on U with Re(f) = u (proof, use the above lemma to find v). Anti-derivatives (primitives): If f is holomorphic on U where U is a rectangle or an open disc, then there is a holomorphic h on U with h = f. (proof, use z the above lemma to find h). Improved version: If f continuous on U and is holomorphic on U \ {p} where U is a rectangle or an open disc, then there is a holomorphic h on U with h z = f. Complex Line (Curve) Integral 3

4 At first sight, complex integration is not really anything different from real integration. For a continuous complex-valued function φ : [a, b] R C, we define b b b φ(t)dt = Reφ(t)dt + i Imφ(t)dt. a a a Let : [a, b] U be a (complex) C curve (or piecewise C ) on U and f be a (complex values) continuous function on U, we define f(z)dz := b a f((t))d(t) = b a f((t)) (t)dt. (Analogy of the fundamental theorem of calculus) Let U C be an open set, and : [a, b] U be piecewise C. If f is holomorphic on U, then f (z)dz = f((b)) f((a)). z This, usually is used together with the existence of the anti-derivative. Another important property is: f(z)dz max z f(z) length(). Cauchy s theorem and Cauchy s integral formula (simple version) Using the fact (see above) about the existence of the anti-derivative (primitive) of holomorphic functions on a disc or rectangle (together with the improved version), we can prove Cauchy s integral theorem. Assume f is holomorphic on an open disc U, and : [a, b] U is a piecewise C curve with (a) = (b) (closed curve). Then f(z)dz = 0. Cauchy s integral formula. Assume f is holomorphic is a neighborhood of D(z 0, r) (i.e. D(z 0, r) U), and f is holomorphic on U. Let = D(z 0, r), i.e. (t) = z 0 + r(cos t + i sin t), t [0, 2π], Then, for each z D(z 0, r), f(z) = 2πi ζ z dζ = 2πi D(z 0,r) 4 ζ z dζ.

5 Remark: The proof of Cauchy s Theorem is based on the result about the existence of primitive of holomorphic functions. The Proof of Cauchy s Integral Formula is based on the fundamental theorem of calculus and the existence of the anti-derivative (primitive) of ( f(z))/(ζ z) (improved version) and the following fundamental result of integral: for z D(z 0, r), dζ =. 2πi D(z 0,r) ζ z Cauchy s theorem and Cauchy s integral formula for more general curves Cauchy s integral formula (and Cauchy s theorem) for more general curves: the above version of Cauchy s integral formula with special curve (i.e. = D(z 0, r)) should have served our purpose for most cases. To get a more general version (for more general curve ), we use the following elementary but very important property for the (line)-integral of holomorphic functions, namely the integral (assuming that f is holomorphic around ) f(z)dz does not change if a small (localized) change is made in the curve. More precisely, if f is holomorphic on U and, µ are two curves in U which coincide except on an interval where both, µ have the images which lie in a disc contained in U. Then f(z)dz = µ f(z)dz. The above observation allows us to stretch from one curve (for example, the circle = D(z 0, r)) to another curve by still keeping the integral value. For example, we can get the following two results: : If f is holomorphic on C \ {0}, then f(z)dz = r f(z) r2 for any 0 < r < r 2, where r is the circle of radius r centered at the origin, or in the annulus case, : If f is holomorphic on the annulus {z r < z < R}, then f(z)dz = r f(z) r2 for any r < r < r 2 < R, where r is the circle of radius r centered at the origin. 5

6 Cauchy s integral formula and theorem (general form): Cauchy s Theorem Let U C be an open set and f is holomorphic on U. Then f(z)dz = 0 for each piecewise C closed curve in U which can be deformed in U through closed curves to a closed curve lying entirely in a disc contained in U. Cauchy s Integral formula For any z U, we have f(z) = 2πi ζ z dζ for any piecewise C closed curve in U \ {z} which can be continuously deformed in U \{z} to a circle of radius r > 0 centered at z (with counter-clockwise orientation) In particular, we have Cauchy s integral formula for an Annulus Let f be holomorphic on D(P, r 2 )\D(P, r ). Then, for any r < s < s 2 < r 2, f(z) = 2πi z P =s 2 ζ z dζ 2πi z P =s ζ z dζ. Basic Properties of Power Series n= a n z n (you should know about convergence, absolute convergence and uniformly convergence): Consider the power series n= a n (z P ) n. WLOG, we assume that P = 0. Abel s Lemma: If n= a n z n converges at some z, then it converges at each w D(0, z ). Remark: This lemma tells us that the domain of convergence is a disc. The r = sup{ z n= a n z n converges} is called the radius of convergence and D(0, r) is the disc of convergence. Absolute and Uniform Conv.: Let n= a n z n be a power series with radius of convergence r. Then, for any 0 R < r, the series converges uniformly and absolutely on D(0, R). 6

7 Hadamard Criterion (root test): Set ρ = lim sup a n /n. Then the radius of convergent is /ρ. This is cooler than the ratio test in that it is more applicable. Theorem. If f(z) = n= a n z n with ρ <, then f is holomorphic on disc {z z < /ρ}, and indeed infinitely (complex) differentiable with Useful formulas: e z = n=0 f (z) = na n z n. z = n=0 z n. z n n!, sin z = ( ) n (2n + )! z2n+, cos z = n=0 n=0 ( ) n (2n)! z2n. Properties of Holomorphic Functions (all derived from Cauchy s integral formula): Cauchy s Formula for derivatives. Let f be holomophic on an set U C. Then f (k) exists on U for all k = 0,, 2,... (so f (k) are holomorphic). Moreover, if D(P, r) U, and z D(P, r), then, for any k = 0,, 2,..., f (k) (z) = k! dζ. 2πi D(P,r) (ζ z) k+ Remark: This is a very surprising result, it tells us that the holomorphic function, which is required by our definition to be continuously differentiable, is actually C! Cauchy s estimate: Let f be holomorphic on D(P, r) U and let M = max z D(P,r) f(z). f (n) (P ) n!m r n. From the proof of Cauchy s Formula for derivatives, we also get that If φ is continuous on D(P, r), then the function defined by is holomorpihc on D(P, r). f(z) = φ(ζ) 2πi D(z 0,r) ζ z dζ 7

8 Morera s Theorem (sort of opposite to Cauchy s integral theorem). Assume that f is continuous on an (connected) set U C. Assume that for every piecewise C curve : [0, ] U with (0) = () (so is closed), f(z)dz = 0. Then f is holomorphic on U. The proof is similar to the existence of antiderivatives for holomorphic functions Remark: () Morera s Theorem provides an alternative way to determine whether f is holomorphic. Goursat s theorem. If f f has complex derivative at each point in U (where U is an open set in C), then f is holomorphic. Remark: The proof uses Morera s Theorem under the observation that, from the proof of Morera s Theorem, one only needs to prove f(z)dz = 0 for any triangular path to conclude that f is holomorphic. Power Series Theorem. Let f be holomorphic on an open set U. Assume that D(P, r) U. Then f can be expanded to a power series on D(P, r), i.e., f(z) = a n (z P ) n, n=0 z D(P, r), where a n = f (n) (P ) n! = dζ. 2πi D(P,r) (ζ P ) n+ Idea of the proof: Use Cauchy s formula f(z) = 2πi fact that, for ζ P = r and z P < r, D(P,r) dζ, and the ζ z ζ z = ζ P z P ζ P = n=0 (z P ) n (ζ P ) n+. Theorem Let f n be a sequence of holomoprphic function f on U. Suppose that {f n } is uniformly convergent on every compact subsets of U. Then its limit f is also holomoprphic function on U. Moreover, the derivatives {f n (k) } is also uniformly convergent to f (k) on every compact subsets of U for every integer k 0. 8

9 Liouville s theorem A bounded holomorphic function f on C is constant. Idea of pf: Show a n = 0 for n by estimating a n = 2πi D(P,R) and then let R goes to. (ζ P ) dζ n+ 2π max z D(P,R) f(z) 2πR, R n+ More general version Assume that f is holomorphic on C and f(z) C z k, for some positive integer k, holds for all z with z >, Then f is a polynomial in z of degree at most k. Fundamental Theorem of Algebra Every non-constant complex polynomial P has a root. Moreover, we can write P (z) = C (z α ) (z α k ). Zeros of Holomorphic Functions(all derived from the power series theorem (the approach by using power series is called Weierstrass method): The tool to study the Zeros of Holomorphic Functions is the following fact (see the power series theorem above), Let f be holomorphic on an open set U. Then f can be expanded to a power series on D(P, r) U, i.e., f(z) = a n (z P ) n, n=0 z D(P, r), where a n = f (n) (P ) n! = dζ. 2πi D(P,r) (ζ P ) n+ Theorem (Principle of Analytic Continuation): Let f be holomorphic on a connected open set U. Assume that there is z 0 U such that f (k) (z 0 ) = 0 for all integer k 0. Then f 0 on U. Remark: The proof goes like this: Let A = {z U f (k) (z) = 0 for all integer k 0}. Claim A is both closed and open (relative to U). So A = U (since U is connected). 9

10 Definition Let z 0 U with f(z 0 ) = 0. Then the least positive integer n such that f (n) (z 0 ) = 0 is called the (zero) order of f at z 0. Remark It is equivalently to say that, around z 0, f(z) = j=n f (j) (z 0 ) (z z 0 ) j. j! Corollary (behavior of f near zero points): Let f be holomorphic on a connected open set U. Let z 0 U with f(z 0 ) = 0. Then there is an integer k 0 such that f(z) = (z z 0 ) k g(z) on U, where k is the (zero) order of f at z 0, and g is a holomorphic function on U with g(z 0 ) 0. In particular, all zeros of f are isolated points. The following fact may be useful (together with above): If g is continuous at z 0 and g(z 0 ) 0, then g has no zeros in a neighborhood of z 0. Corollary (Uniqueness Theorem). Let f be holomorphic on an (connected) open set U. Assume that {z j } {z f(z) = 0} (i.e. f(z j ) = 0), and z j z 0 U, then f 0. Corollary (Uniqueness Theorem). Let f, g be holomorphic on an (connected) open set U. Assume that {z j } {z f(z) = g(z)} (i.e. f(z j ) = g(z j )), and z j z 0 U, then f g. Corollary (Uniqueness Theorem). Let f, g be holomorphic on an (connected) open set U. Assume that f g = 0 on U. Then either f 0 or g 0 on U. Theorem(Counting the number of zeros). If f 0 is holomorphic on D(P, r) U and has zeros z,..., z k of orders n,..., n k in D(P, r). Then f (ζ) k 2π D(P,r) dζ = j= n j. Idea of pf: Write k f(z) = (z z j ) n j H(z) j= where H is a holomorphic function on U without zeros on D(P, r), and then compute f (z)/f(z). The theorem then follows from the Cauchy s theorem and Cauchy s integral formula. 0

11 More General Version. If f 0 is holomorphic on D(P, r) U and has zeros z,..., z k of orders n,..., n k in D(P, r). Let g be a holomorphic function on U. Then g(ζ) f (ζ) k 2π D(P,r) dζ = n j g(z j ). Rouche s Theorem. Suppose that f, g are holomorphic on D(P, r) U. Assume that for each ζ D(P, r), g(ζ) < + g(ζ). Then j= f (ζ) 2π D(P,r) dζ = g (ζ) 2π D(P,r) g(ζ) dζ. That is the number of zeros of f in D(P, r) is the same as the number of zeros of g in D(P, r), counting multiplicities. Remark: Rouche s Theorem is often stated with the stronger hypothesis: each ζ D(P, r), g(ζ) < g(ζ). This gives the following corollary: Useful version of Rouche s Theorem Suppose that f, g are holomorphic on D(P, r) U. Assume that for each ζ D(P, r), h(ζ) < g(ζ). Then the functions g(z), g(z)±h have the same number of zeros in D(P, r), counting multiplicities. The above facts about zeros of f can also be used to study a-points of f, i.e. the zeros of f(z) a. This leads to study the behavior of f as a mapping. The Open Mapping Theorem. If f : U C is a non-constant holomorphic function, where U is a connected open set. Then f(u) is also open. Idea of pf: A trivial corollary of the theorem below. Theorem about pre-images. If f : U C is a non-constant holomorphic function, where U is a connected open set. Assume that z 0 U with f(z 0 ) = w 0 with order k (i.e. f(z 0 ) w 0 = (z z 0 ) k g(z) with g being holomorphic on U and g(z 0 ) 0). Then there are δ, ρ > 0 such that each w D(w 0, ρ)\{w 0 } has exactly k distinct pre-images in D(z 0, δ) and each preimage is a simple point of f. In particular, if in addition we assume that f (z 0 ) 0, then every w D(w 0, ρ)\{w 0 } has a unique preimage inside D(z 0, δ). Remark: This theorem gives a more precise description than the Open Mapping Theorem. Idea of pf: Since zeros of holomorphic functions are isolated, we can choose δ > 0 such that f(z) w 0 and f (z) have no zeros on D(z 0, δ)\{z 0 }. Let

12 ρ = min z D(z0,δ) f(z) w 0. Then, for every w D(w 0, ρ) (i.e. w w 0 < ρ), on z D(z 0, δ), w w 0 < ρ f(z) w 0. Rouche s theorem then implies that k = # of zeros of f(z) w 0, C.M., in D(z 0, δ) = # of zeros of f(z) w, C.M., D(z 0, δ). Moreover, since f (z) have no zeros on D(z 0, δ)\{z 0 }, each preimage is a simple point of f. If in addition we assume that f (z 0 ) 0, then k =. So every w D(w 0, ρ)\{w 0 } has a unique preimage inside D(z 0, δ). Theorem about inverse. If f : U C is a non-constant holomorphic function, where U is a connected open set. Assume that z 0 U such that f (z 0 ) 0. Let δ, ρ > 0 be as above, and V = f (D(w 0, ρ)) D(z 0, δ) (which is open). Then f V : V D(w 0, ρ) is one to one and onto. Furthermore, for w D(w 0, ρ), f (w) = ζf (ζ) 2πi D(z 0,δ) w dζ. Idea of pf: The fact that f is one to one and onto is proved above. To show the formula (for f (w)), we use the result stated above that, with h(z) = f(z) w and g(z) = z, g(ζ) h (ζ) k 2π D(P,r) h(ζ) dζ = n j h(z j ), where z j are zeros of h. The Maximum modulus principle. Suppose that f : U C is a non-constant holomorphic function, where U is a connected open set. If there is a point P U such that f(p ) f(z) for all z U (i.e. f(p ) is maximal). Then f is constant. j= Idea of pf: Consequence of the open mapping theorem. Schwarz Lemma. Let f be holomorphic on the unit disc. Assume that f(z) for all z (i.e. f maps from the unit-disc to the unit-disc) and f(0) = 0. Then f(z) z and f (0). Laurent Series for Holomprphic Functions on an Annulus:. Consider the Laurent Series n= a n (z P ) n. 2

13 It is convergent if and only if the both two series n=0 a n (z P ) n (this is a power series) and n=0 a n [(z P ) ] n (this is not a power series, but is similar to a power series in terms of the properties). From what we know about power series (see above), we know that the domain of convergence is r < z P < r 2, where 0 r < r 2, (since the domain of convergence of the seris n=0 a n (z P ) n is z P < r 2 and the domain of convergence of the seris n=0 a n [(z P ) ] n is z P > r ). Furthermore, we have Compare with the Abel s lemma: If the Laurent Series n= a n (z P ) n n= a n (z P ) n converges at two points z, z 2 (not equal to P ), and if z P < z 2 P, then it converges on z P < z P < z 2 P. Compare to Absolute and Uniform Conv.: Let n= a n (z P ) n be convergent on r < z P < r 2. Then, for any r < r < r 2 < r 2, it converges uniformly and absolutely on r < z P < r 2. Uniquness of Laurent sereis: Let n= a n (z P ) n be convergent on D(P, r 2 )\D(P, r ), where 0 r < r 2 to a function f. Then for any r < r < r 2 and each integer j a j = 2πi D(P,r) dζ. (ζ P ) j+ Expansion of a Laurent sereis (compare the Expansion of power series for holo. func): Let f be holomorphic on an annular domain D(P, r 2 )\D(P, r ). Then we have Laurent s expansion of f: f(z) = j= a j (z P ) j, where a j = 2πi z P =r (ζ z) dζ, for all r j+ < r < r 2. Idea of pf: Similar to the proof of the Expansion of power series for holo. functions, by applying the Cauchy s integral formula for an annulus. Holomprphic Functions with Isolated Singularities (uisng the Laurent series on the punctured disk): 3

14 Let f be holomorphic on U\{P } where U C is open and P U. From above, we can write, for all z D(P, r)\{p } with D(P, r) U, f(z) = j= a j (z P ) j, where a j = dζ, for all 0 < s < r, 2πi z P =s (ζ z) j+ Definition Such point P is called an isolated Singularity of f. Riemann s Extension Theorem. If f is holomorphic on D(P, r 0 )\{P } (where r 0 > 0) and bounded. Then f can be extended holomorphically across the point P, i.e. there is a holomorphic function f on U such that f U\{P } = f. Idea of pf: Consider g(z) := (z P ) 2 f(z), and then use the power series expansion argument. Theorem. There are three mutually exclusive possibilities for an isolated Singularity P of f: (i) f is bounded on U\{P } or equiv. a j = 0 for j < 0. In this case, we call P a removable singularity. Example: f(z) = cos z z. (ii) lim z P f(z) = or equiv. a j = 0 for all j < k for some positive integer k. Example: f(z) = /z. (iii) Neither (i) nor (ii). In this case, we call P an essential sigularity. Example: f(z) = e /z. Idea of pfs of the equivalence in (i) and (ii): (i) = (Riemann s extention theorem). = because f is indeed represented by a power series on D(P, r 0 ) is it is holomorphic on D(P, r 0 ), thus f is bounded on D(P, r 0 ). (ii): = : Since lim z P f(z) =, f is never zero on D(P, s) for some s < r 0. Consider g = f and use Riemann s extention theorem, we get f = (z a)m Q(z) on where D(P, s)\{p }, hence f(z) = j= m b j (z P ) j. =: Assume that f(z) = j= k a j (z P ) j with k > 0 being fixed and a k 0, it is easy to see that lim z P f(z) = by noticing that f(z) = z P k ( a k j= k+ a j (z P ) j+k ). 4

15 Theorem about the essential singularity (Casorati-Weierstrass Theorem). If f is holomorphic on D(P, r 0 )\{P } and P is an essential sigularity. Then, for every 0 < r < r 0, f(d(p, r)\{p }) is dense in C. Idea of pf: Assume that a C with f(z) a > ɛ on D(P, r)\{p }. Consider g(z) = /(f(z) a) and use Riemann s Extension Theorem there is ĝ which is holomorphic on D(P, r) with ĝ(z) = g(z) for all z D(P, r)\{p }. So f(z) = a +. This means either P is a removable singularity or a pole, which leads ĝ(z) a contradiction. Theorem about the pole. Assume that P is a pole of f. We can write D(P, r)\{p }, f(z) = j= k a j (z P ) j with a k 0. We say that P os a pole of order k. We have: (i) On D(P, r)\{p }, f(z) = (z P ) k g(z) where g is holomorphic and g(p ) 0. For j = k, k +, k + 2,..., (ii) a j = (k + j)! ( z )k+j ((z P ) k f) P. Remark () The formula in (ii) is very convenient to use to compute the (coefficients of) the Laurent expansion. (2) The term a is expecially important, which is called the residue of f at the point P, denoted by Res P (f). Definition Let U be an open set and P U. Let be a piecewise C closed curve in U such that P is not on the curve. We define the index of with respect to P, denoted by ind (P ) as ind (P ) := 2πi ζ P dζ. Remark: ind (P ) is always an integer. The proof is as follows: Note that ind (P ) = 2πi ζ P dζ = b (t) 2πi a (t) P dt, and try to show that exp( b a (t) (t) P dt) = by proving that b (s) g(t) := ((t) P ) exp( a (s) P ds) is a constant, so g(a) = g(b) gives exp( b a 5 (t) dt) =. (t) P

16 Definition An open set U is is called holomorphically simply connected if if U is connected, and if every holomorphic function on U has ab anti-derivative (for example, the open discs or rectangles, etc.). Lemma U is h.s.c. if and only if for each holomorphic function f on U and each piecewise C closed curve in U, f = 0. Residue Theorem Let U C be an h.s.c. and P,..., p k U be distinct. Assume that f us holomorpihc on U\{P,..., P k }. Then, for any closed, piecewise C curve in U\{P,..., P k }, k f(z)dz = 2πi Res f (P j ) Ind (P j ). j= Generalized Cauchy s theorem Suppose that f is analytic in an open disk and is a closed curve in. Assume that a. Then Proof: Use Res f(z) (a) = f(a). z a Ind (a)f(a) = f(z)dz 2πi z a. Argument Principle (Counting the number of zeros and poles). If f 0 is meremorphic (i.e. holomorphic except poles) on D(P, r) U and has zeros a,..., a k of orders n,..., n k in D(P, r) and poles b,..., b l of orders m,..., m l in D(P, r). Then f (ζ) k l 2π D(P,r) dζ = n j m t. Idea of pf: Apply the Residue Theorem by noticing that Res f (a j ) = n j and f Res f (b t ) = m t. f Residue Calculation formula: Let P be a ploe of f order k, then a = j= t= (k )! ( z )k ((z P ) k f) P. 6

17 Application of the Residue Theorem in calculation of definite integrals: Type : + P (x) f(x)dx where f(x) = Q(x) 4.6. on Page 28. with deg P deg Q 2 see Example Type 2 (Fourier-type integral): + f(x)eiαx dx with α > 0 with lim z,imz 0 f(z) = 0. See Example on Page 30 and Example on Page 3. This would give + cos x dx, and + +x 2 sin x x dx. Type 3: + 0 R(x)dx, where R is a rational function. We assume that R(z) = P (z)/q(z) with no pole on the positive real axis and degree deg P deg Q 2. The trick is to use an auxiliary function f(z) := R(z) log z on the domain C\[0, + ]. See Example on Page 35. Type 4: Integrals + 0 R(x)x α with R(z) = P (z)/q(z) and deg P +α < deg Q. See Example on Page 33. Holomorphic Functions as Mappings ( Riemann s View): If f is holomorphic around P, then f is conformal at P (i.e. it preserves the angles and stretches in all directions (see Theorem on Page 40). If f : U V is holomorphic (non-constant), U is open, and V = f(u), then V is also open (by the open mapping thoerem). f : U V is said to be conformal or biholomoprhic if f is holomorphic, oneto-one and onto. Examples of conformal mappings:. f : C C is conformal if and only if f is linear. 2. f : D(0, ) D(0, ) is conformal if and only if f is a Mobius transformation (a consequence of Schwarz-Pick Lemma). 3. f : {z Imz > 0} D(0, ) is conformal if and only if f(z) = r iθ z a, Ima > 0. z ā 4. f : {z Imz > 0} {z Imz > 0} is conformal if and only if f(z) = az + b cz + d, where a, b, c, d are real numbers, and ad bc > 0. 7

18 Normal Families: Let F = {f α } α Λ be a family of holomorphic functions on U C. We say that F is normal family if every sequence {f n } F has a sequence that converges uniformly on every compact subsets of U. Montel s Theorem Let F = {f α } α Λ be a family of holomorphic functions on U C that is bounded on compact subsets, i.e. for every compact set K U, there is a M K > 0 such that Then F is normal. f(z) M k, z K, f F. Riemann Mapping Theorem (Analytic Version) If U is a holomorphically simply connected open set in C and U C, then U is conformally equivalent to the unit disc. Idea of pf: Set, for a fixed P U,. F = {f f : U D(0, ) is holomorphic, and one-to-one, f(p ) = 0}. 2. Use the normality to show that there is f 0 F such that f 0(P ) = sup f (P ). f F 3. Show that f 0 maps U onto the unit disc D(0, ). If U is simply connected, then U is a holomorphically simply connected. So the Riemann Mapping Theorem: If U is simply connected and U C. Then U is conformally equivalent to the unit disc. So the beauty of this theorem is that it basically says that the topological property of U implies the analytic property of U. Little Picard Theorem (Generalization of Liouville s theorem) If f is entire and omits 2 values, then it is constant. Idea of pf: Step. From omitting 2 values to omitting a lot of values (from spotting 2 cockroaches to a lot of cockroaches), this can be achieved by proving the following lemma: If U is simply connected and f is holomorphic on U and does not assume the values of 0 and. Then there is a holomorphic curve on U such that f(z) = exp(iπ cosh[2g(z)]). 8

19 Furthermore, it can be verified that g does not assume, on U, the values of π log( n+ n )+ mπi for positive integer n and all integer m. Since these 2 points from a vertices of a grid of rectangles with diagonal < 2, we have that g(u) contains no disc of radius. Step 2. We prove the so-called Bloch s theoremlet f be analytic on a region containing D(0, ) and f(0) = 0, f (0) =. Then there f(d(0, )) constains a disk of radius /72 (which is independent of f!). Step 3. The proof of little picard theorem can thus be proved by considering g(z + z 0 )/g (z 0 ) (by choosing some z 0 with g (z 0 ) 0 and by Bloch s theorem, g(d(z 0, R)) contains a disc of redius (/72)R, which contains a disc of radius when R is bigger enough, which contradicts with that fact that g(u) contains no disc of radius derived in step. The proof of Bloch s theorem: We first prove the following lemma: LemmaLet f be holomorphic on D(0, ) with f(0) = 0, f (0) =. Assume that f(z) M for all z D(0, ). Then M and f(d(0, )) D(0, /(6M)). id of proof: Let 0 < r < and f(z) = z + a 2 z 2 +. According to Cauchy s theorem, a n M/r n for n. Yence = a M and a n M.. So we can prove that for z = (4M), f(z) (6M). Thus by Rouche s theorem, since for every w D(0, /6M), w f(z) on z = (4M), f(z) and f(z) w have the same number of zeros inside z (4M). But f(0) = 0, so there exists z 0 with f(z 0 ) = w. This proves the lemma. Corollary Let g be holomorphic on D(0, R) with g(0) = 0, g (0) = µ. Assume that g(z) M for all z D(0, R). Then g(d(0, R)) D(0, R 2 µ 2 /(6M)). So to apply Corollary to proof Bloch s theorem, the main task is to drop the condition that f(z) M. Here is the trick: Let K(r) = max{ f (z) : z = r} and consider h(r) = ( r)k(r). Then h is continuous and h(0) =, h() = 0. Let r 0 = sup{r : h(r) = }. Then h(r 0 ) = and h(r) for r < r 0 (so we get a bound for K(r) thus the derivative f (thus the value of f)). Here is how to do it: Take a with a = r 0 and f (a) = K(r 0 ). Then f (a) = ( r 0 ). 9

20 Now for z a 2 ( r 0) := ρ 0 (so z 2 ( + r 0), from the definition of r 0, f (z) K( 2 ( + r 0) = h( 2 ( + r 0)[ 2 ( + r 0)] < [ 2 ( + r 0)] = /ρ 0. Consider g(z) = f(z +a) f(a). Notice that for z D(0, 2 ρ 0), the line segment [a, z + a] lies in D(a, ρ 0 ). Hence g(z) f (w)dw z < /2. ρ 0 Corollary implies that g(d(0, 2 ρ 0)) D(0, σ where σ = ( ρ ) 2 ( 3 0 ρ 2 2 0) 6 ( ) = Marty s criterion for nromality: Let F be a family of holomrphic functions on a region U of C. Then F is normal (in the extended sense) if and only if for every compact subset K of U there is a constant C K such that all z K and f F, where f # (z) := f # (z) C K f (z) + f(z) 2. Zalcman s theroem. Let F be a family of holomrphic functions on the unit-disk D of C. Assume that F is NOT normal (in the extended sense). Then there exists real number r with 0 < r <, a seqeucne of points z n with z n < r, a sequence {f n } F, and a seqeunce of positive numbers ρ n with lim n + ρ n 0 such that f n (z n + ρ n ξ) g(ξ) uniformly on every compact subsets K C, and g is not constant. Montel s (general) Theorem Let F be a family of holomrphic functions on a region U of C. Assume that for every f F, f(z) 0, for all z U. Then F is normal on U. The proof uses the Zalcman s theorem and the little Picard s theroem. 20

21 Big Picard s theorem Let f be holomorphic on D(P, r)\{p } which has an essential singularity at P. Then for every 0 < s < r, the image f(d(p, s)\{p }) covers the whole complex plane, except possibly for one complex value. Proof. Assume the theorem is false, WLOG, we can assume that f(d(0, )\{0}) omits the value of 0 and. We claim that 0 is not an essential singluarity which contradicts with our assumption. To show that claim, consider f n (z) = f(z/n), 0 < z <. Then, from the (generalized) Montel s theorem, {f n } is normal on 0 < z <. Thus there exists a subsequence {f nk } which is uniformly convergent on every compact subset of 0 < z < or or uniformly divergent every compact subset of 0 < z <. In the first case, i.e. {f nk } is uniformly convergent on every compact subset of 0 < z < implies that {f nk } is uniformly bounded on z = /2, i.e. there exists M > 0 such that f nk (z) M for all z = /2. This means that M for all ζ = /(2n k ). Thus, by the maximal principle, M on {ζ 0 < ζ < /2}. Hence, by the Riemann s extension theorem, 0 is a removable singularity. In the second case, a similar argument applies to /f shows that f tends to when z 0. This means that 0 is a pole for f. Thus the claim is proved. 2