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Complex Analysis University of Cambridge Mathematics Tripos Part IB Complex Analysis Lent, 2018 Lectures by G. P. Paternain Notes by Qiangru Kuang Contents Contents 1 Basic notions 2 1.1 Complex differentiation ....................... 2 1.2 Power series .............................. 4 1.3 Conformal maps ........................... 8 2 Complex Integration I 10 2.1 Integration along curves ....................... 10 2.2 Cauchy’s Theorem, weak version .................. 15 2.3 Cauchy Integral Formula, weak version ............... 17 2.4 Application of Cauchy Integration Formula ............ 18 2.5 Uniform limits of holomorphic functions .............. 20 2.6 Zeros of holomorphic functions ................... 22 2.7 Analytic continuation ........................ 22 3 Complex Integration II 24 3.1 Winding number ........................... 24 3.2 General form of Cauchy’s theorem ................. 26 4 Laurent expansion, Singularities and the Residue theorem 29 4.1 Laurent expansion .......................... 29 4.2 Isolated singularities ......................... 30 4.3 Application and techniques of integration ............. 34 5 The Argument principle, Local degree, Open mapping theorem & Rouché’s theorem 38 Index 41 1 1 Basic notions 1 Basic notions Some preliminary notations/definitions: Notation. • 퐷(푎, 푟) is the open disc of radius 푟 > 0 and centred at 푎 ∈ C. • 푈 ⊆ C is open if for any 푎 ∈ 푈, there exists 휀 > 0 such that 퐷(푎, 휀) ⊆ 푈. • A curve is a continuous map from a closed interval 휑 ∶ [푎, 푏] → C. It is continuously differentiable, i.e. 퐶1, if 휑′ exists and is continuous on [푎, 푏]. • An open set 푈 ⊆ C is path-connected if for every 푧, 푤 ∈ 푈 there exists a curve 휑 ∶ [0, 1] → 푈 with endpoints 푧, 푤. Definition (Domain). A domain is an non-empty path-connected open subset of C. 1.1 Complex differentiation The goal of this course is to study functions 푓 ∶ 푈 → C where 푈 ⊆ C is open or is a domain. Given such an 푓, we may write 푓(푥 + 푖푦) = 푢(푥, 푦) + 푖푣(푥, 푦) where 푢, 푣 ∶ 푈 → R are the real and imaginary part of 푓. Here we use (푥, 푦) ∈ R2 to denote the coordinates of 푎 + 푏푖 ∈ C. Definition (Differentiable, holomorphic). 1. 푓 ∶ 푈 → C is differentiable at 푤 ∈ 푈 if the limit 푓(푧) − 푓(푤) 푓′(푤) = lim 푧→푤 푧 − 푤 exists. 푓′(푤) is called the derivative of 푓 at 푤. 2. 푓 is holomorphic at 푤 if there exists 휀 > 0 such that 푓 is differentiable at all points of 퐷(푤, 휀). 푓 is holomorphic on 푈 if it is differentiable at all 푤 ∈ 푈. Equivalent, 푓 is holomorphic at at all 푤 ∈ 푈. Remark. 1. There is an alternative term analytic. In actuality, it is the same as holomorphic for complex functions. However, it comes with a flavour associated with the Taylor expansion and sometimes defined in terms of such. Later in the course we will prove that the definitions are equivalent. 2. Complex differentiation follows the same rules as real differentiation. For example, sums of differentiable functions are differentiable, product, quo- tient, chain rules etc also hold. 2 1 Basic notions Definition (Entire). An entire function is a holomorphic function 푓 ∶ C → C. Example. 1. Polynomials are entire functions. 푝 2. If 푝(푧) and 푞(푧) are polynomials with 푞(푧) not identically zero, then 푞 is holomorphic on C \{zeros of 푞}. In IB Analysis II we studied function from R푛 to R푚 and their differentiability. Indeed a complex function C → C can be view as a function R2 → R2 so how does complex differentiability relates to differentiablitiy in R2? It turns out that in addition to satisfy the differentiability on R2, the function has to satisfy a particular partial differential equation. Recall that 푢 is differentiable at (푐, 푑) ∈ 푈 if there exists (휆, 휇) ∈ R2 such that 푢(푥, 푦) − 푢(푐, 푑) − (휆(푥 − 푐) + 휇(푦 − 푑)) → 0 √(푥 − 푐)2 + (푦 − 푑)2 as (푥, 푦) → (푐, 푑). In this case 풟푢(푐, 푑) = (휆, 휇) is the derivative of 푢 at (푐, 푑). If this holds then 휆 = 푢푥(푐, 푑) and 휇 = 푢푦(푐, 푑), the partial derivatives of 푢 at (푐, 푑). Theorem 1.1 (Cauchy-Riemann equations). 푓 ∶ 푈 → C is differentiable at 푤 = 푐 + 푖푑 ∈ 푈 if and only if the functions 푢 and 푣 are differentiable at (푐, 푑) and 푢푥(푐, 푑) = 푣푦(푐, 푑) 푢푦(푐, 푑) = −푣푥(푐, 푑) in which case ′ 푓 (푤) = 푢푥(푐, 푑) + 푖푣푥(푐, 푑). Proof. From the definition, 푓 will be differentiable at 푤 with derivative 푓′(푤) = 푝 + 푖푞 if and only if 푓(푧) − 푓(푤) − 푓′(푤)(푧 − 푤) lim = 0 푧→푤 |푧 − 푤| or equivalently, splitting into real and imaginary parts, 푢(푥, 푦) − 푢(푐, 푑) − (푝(푥 − 푐) − 푞(푦 − 푑)) lim = 0 (푥,푦)→(푐,푑) √(푥 − 푐)2 + (푦 − 푑)2 푣(푥, 푦) − 푣(푐, 푑) − (푞(푥 − 푐) + 푝(푦 − 푑)) lim = 0 (푥,푦)→(푐,푑) √(푥 − 푐)2 + (푦 − 푑)2 since 푓′(푤)(푧 − 푤) = (푝(푥 − 푐) − 푞(푦 − 푑)) + 푖(푞(푥 − 푐) + 푝(푦 − 푑)). So 푓 is differentiable at 푤 with derivative 푓′(푤) = 푝 + 푖푞 if and only if 푢 and 푣 are differentiable at (푐, 푑) with 풟푢(푐, 푑) = (푝, −푞) 풟푣(푐, 푑) = (푞, 푝) hence the result. 3 1 Basic notions Example. Let 푓(푧) = 푧. Then 푓(푥 + 푖푦) = 푥 − 푖푦, 푢(푥, 푦) = 푥, 푣(푥, 푦) = −푦. We have 푢푥 = 1 ≠ −1 = 푣푦 so 푓 is not differentiable anywhere in C. Remark. 1. We could have discovered Cauchy-Riemann as follows: let 푧 = 푤 + ℎ where ℎ ∈ R. Then ′ 푓(푤 + ℎ) − 푓(푤) 푓 (푤) = lim = 푢푥(푐, 푑) + 푖푣푥(푐, 푑). ℎ→0 ℎ Let 푧 = 푤 + 푖ℎ, we get ′ 푓(푤 + 푖ℎ) − 푓(푤) 푓 (푤) = lim = 푣푦(푐, 푑) − 푖푢푦(푐, 푑). ℎ→0 ℎ As 푓 is assumed to be differentiable, these two must agree. 2. Later on we’ll see that if 푓 is holomorphic, so is 푓′. This will imply right away that all partial derivatives of 푢 and 푣 exist and are continuous, i.e. 퐶∞. Thus 푢푥푥 = 푣푦푥 푢푦푦 = −푣푥푦 By symmetry of second derivatives, we get that 푢 satisfies the Laplace equation 푢푥푥 + 푢푦푦 = 0. Hence 푓 is holomorphic implies that the real and imaginary parts are harmonic functions. Corollary 1.2. Let 푓 = 푢+푖푣 ∶ 푈 → C. Suppose the functions 푢 and 푣 have continuous partial derivatives everywhere in 푈 and satisfy Cauchy-Riemann equations. Then 푓 is holomorphic in 푈. Proof. From IB Analysis II, 푢 and 푣 are differentiable. The result follows from the previous theorem. Corollary 1.3. Let 푓 ∶ 퐷 → C be holomorphic on a domain 퐷 and suppose 푓′(푧) = 0 for all 푧 ∈ 퐷. Then 푓 is constant on 퐷. Proof. Follows from the analogous result for differentiable functions on a path- connected subset of R2 (Mean Value Inequality from IB Analysis II). 1.2 Power series Recall that 4 1 Basic notions Theorem 1.4 (Radius of convergence). Let 푐푛 be a sequence of complex numbers. Then there exists a unique 푅 ∈ [0, ∞], the radius of convergence of the series, such that ∞ 푛 ∑ 푐푛(푧 − 푎) , 푧, 푎 ∈ C 푛=0 converges absolutely if |푧 − 푎| < 푅 and diverges if |푧 − 푎| > 푅. If 0 < 푟 < 푅 then the series converges uniformly on {|푧−푎| ≤ 푟}. The radius of convergence is given by 푛 푅 = sup{푟 ≥ 0 ∶ |푐푛|푟 → 0}. 푓(푧) = ∑∞ 푐 (푧 − 푎)푛 Theorem 1.5. Let 푛=0 푛 be a complex power series with radius of convergence 푅 > 0. Then 1. 푓 is holomorphic on 퐷(푎, 푅), ∑∞ 푛푐 (푧 − 푎)푛−1 2. its derivative is given by the series 푛=1 푛 , which also has radius of convergence 푅, (푛) 3. 푓 has derivatives of all orders on 퐷(푎, 푅) and 푓 (푎) = 푛!푐푛, 4. if 푓 vanishes identically on some disc 퐷(푎, 휀) then 푐푛 = 0 for all 푛. 푎 = 0 ∑∞ 푛푐 푧푛−1 Proof. wlog assume . Claim that 푛=1 푛 has radius of convergence 푅. ′ Proof. |푛푐푛| ≥ |푐푛| so its radius of convergence 푅 is at most 푅. 푛 Suppose 0 ≤ 푟 < 푅 and pick 휌 ∈ (푟, 푅). Then ∑ |푐푛|휌 converges and 푛−1 푛 푛|푐푛|푟 푛 푟 푛 = ( ) → 0 |푐푛|휌 푟 ⏟휌 <1 푛−1 ′ so by comparison test ∑ 푛푐푛푟 converges and hence 푅 = 푅. To show the derivative is of the desired form we shall do something clever. Consider the continous function 푛−1 푧푛−푤푛 푗 푛−1−푗 푧−푤 if 푧 ≠ 푤 ℎ푛(푧, 푤) = ∑ 푧 푤 = { 푛−1 푗=0 푛푤 if 푧 = 푤 for 푛 ≥ 1. Consider the series ∞ ∑ 푐푛ℎ푛(푧, 푤). (∗) 푛=1 Claim that for every 푟 < 푅, (∗) converges uniformly on the set {(푧, 푤) ∶ |푧|, |푤| ≤ 푟}. 푛−1 Proof. |푐푛ℎ푛(푧, 푤)| ≤ |푐푛|푛푟 = 푀푛. Since we know ∑ 푀푛 < ∞, by Weier- strass M-test (∗) converges uniformly. 5 1 Basic notions Thus it converges to a continuous function 푔(푧, 푤). By definition ∞ 푛 푛 ∑ 푐 푧 −푤 = 푓(푧)−푓(푤) if 푧 ≠ 푤 푔(푧, 푤) = { 푛=1 푛 푧−푤 푧−푤 ∑∞ 푐 푛푤푛−1 푧 = 푤 푛=1 푛 if As 푔 is continuous, fixing 푤 and letting 푧 → 푤 we get ∞ 푓(푧) − 푓(푤) 푛−1 lim = ∑ 푛푐푛푤 . 푧→푤 푧 − 푤 푛=1 So 푓′(푤) exists and equals to 푔(푤, 푤) as desired. This proves (1) and (2). (3) follows by induction on 푛. Finally if 푓 vanishes identically on a disc about 푎 (푛) then 푓 = 0 for all 푛 and by (3) 푐푛 = 0 for all 푛.
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