stein complex analysis 答案 复分析答案.pdf

SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI ROBERT C. RHOADES Abstract. This contains the solutions or hints to many of the exercises from the Complex Analysis book by Elias Stein and Rami Shakarchi. I worked these problems during the Spring of 2006 while I was taking a Complex Analysis course taught by Andreas Seeger at the University of Wisconsin - Madison. I am grateful to him for his wonderful lectures and helpful conversations about some of the problems discussed below. Contents 1.Chapter 1. Preliminaries to Complex Analysis2 2.Chapter 2. Cauchy’s Theorem and Its Applications8 3.Chapter 3. Meromorphic Functions and the Logarithm9 4.Chapter 4. The Fourier Transform10 5.Chapter 5: Entire Functions11 6.Chapter 6. The Gamma and Zeta Functions13 7.Chapter 7: The Zeta Function and Prime Number Theorem17 8.Chapter 8: Conformal Mappings20 9.Chapter 9: An Introduction to Elliptic Functions23 10.Chapter 10: Applications of Theta Functions25 Date: September 5, 2006. The author is thankful for an NSF graduate research fellowship and a National Physical Science Consortium graduate fellowship supported by the NSA. 1 2ROBERT C. RHOADES 1. Chapter 1. Preliminaries to Complex Analysis Exercise 1. Describe geometrically the sets of points z in the complex plane defi ned by the following relations: (1) |z − z1| = |z − z2| where z1,z2∈ C. (2) 1/z = z. (3) Re(z) = 3. (4) Re(z) c, (resp., ≥ c) where c ∈ R. (5) Re(az + b) 0 where a,b ∈ C. (6) |z| = Re(z) + 1. (7) Im(z) = c with c ∈ R. Solution 1. (1) It is the line in the complex plane consisting of all points that are an equal distance from both z1and z2. Equivalently the perpendicular bisector of the segment between z1and z2 in the complex plane. (2) It is the unit circle. (3) It is the line where all the numbers on the line have real part equal to 3. (4) In the fi rst case it is the open half plane with all numbers with real part greater than c. In the second case it is the closed half plane with the same condition. (5) (6) Calculate |z|2= x2+ y2= (x + 1)2= x2+ 2x + 1. So we are left with y2= 2x + 1. Thus the complex numbers defi ned by this relation is a parabola opening to the “right”. (7) This is a line. Exercise 2.Let h·,·i denote the usual inner product in R2. In other words, if Z = (z1,y1) and W = (x2,y2), then hZ,Wi = x1x2+ y1y2. Similarly, we may defi ne a Hermitian inner product (·,·) in C by (z,w) = zw. The term Hermitian is used to describe the fact that (·,·) is not symmetric, but rather satisfi es the relation (z,w) = (w,z) for all z,w ∈ C. Show that hz,wi = 1 2 [(z,w) + (w,z)] = Re(z,w), where we use the usual identifi cation z = x + iy ∈ C with (x,y) ∈ R2. Solution 2. This is a straightforward calculation 1 2 [(z,w) + (w,z)] =1 2 (zw + wz) = Re(z,w)Re(zw) =1 2 ((z1+ z2i)(w1+ iw2) + (w1+ iw2)(z1− iz2)) = z1w1+ z2w2. Exercise 3.With ω = seiφ, where s ≥ 0 and φ ∈ R, solve the equation zn= ω in C where n is a natural number. How many solutions are there? SOLUTIONS/HINTS TO THE EXERCISES FROM COMPLEX ANALYSIS BY STEIN AND SHAKARCHI3 Solution 3.zn= seiφimplies that z = s 1 nei( φ n+ 2πik n ), where k = 0,1,··· ,n − 1 and s 1 nis the real nth root of the positive number s. There are n solutions as there should be since we are fi nding the roots of a degree n polynomial in the algebraically closed fi eld C. Exercise 4. Show that it is impossible to defi ne a total ordering on C. In other words, one cannot fi nd a relation ? between complex numbers so that: (1) For any two complex numbers z,w, one and only one of the following is true: z ? w, w ? z or z = w. (2) For all z1,z2,z3∈ C the relation z1? z2implies z1+ z3? z2+ z3. (3) Moreover, for all z1,z2,z3∈ C with z3? 0, then z1? z2implies z1z3? z2z3. Solution 4. Suppose, for a contradiction, that i ? 0, then −1 = i · i ? 0 · i = 0. Now we get −i ? −1 · i ? 0. Therefore i − i ? i + 0 = i. But this contradicts our assumption. We obtain a similar situation in the case 0 ? i. So we must have i = 0. But then for all z ∈ C we have z · i = z · 0 = 0 Repeating we have z = 0 for all z ∈ C. So this relation would give a trivial total ordering. Exercise 5.A set Ω is said to be pathwise connected if any two points in Ω can be joined by a (piecewise-smooth) curve entirely contained in Ω. The purpose of this exercise is to prove that an open set Ω is pathwise connected if and only if Ω is connected. (1) Suppose fi rst that Ω is open and pathwise connected, and that it can be written as Ω = Ω1∪ Ω2where Ω1and Ω2are disjoint non-empty open sets. Choose two points ω1∈ Ω1 and ω2∈ Ω2and let γ denote a curve in Ω joining ω1to ω2. Consider a parametrization z : [0,1] → Ω of this curve with z(0) = ω1and z(1) = ω2, and let t∗= sup 0≤t≤1{t : z(s) ∈ Ω 1for all 0 ≤ s δ for all ? 0. But this is a contradiction since z is smooth. Defi ne Ω1and Ω2as in the problem. First to see that Ω1is open let z ∈ Ω1. Then since Ω is open and z ∈ Ω we know that there exists a ball B(z,δ ⊂ Ω. We claim that this ball is actually。