MATH5685 Assignment 3 Due: Wednesday 3 October 1. The open unit disk is denoted D. Q1. Suppose that a n for all n. Show that (1 + a n) converges if and only if a n converges. [Hint: prove that ( N (1 + a N ) n) exp a n.] Q. Where does f(z) = Q3. Show that f(z) = ( ) 1 + z n converge? And what does it converge to? n= e t t z 1 dt is analytic on the right half plane. n 1 ( ) k Q4. For n = 1,, 3,..., find a formula for Γ. (Step 1 is to find a formula; n k=1 Step is to prove that it hold for all n.) Q5. Let f(z) = z (1 + z) (a) Find the image of D under the map f. (b) Find the image of the right half-plane under the map f. Q6. Find a bijective conformal maps which map (a) the square onto the half-disk S = {x + iy : < x < 1, < y < 1} H = {re iθ : < r < 1, < θ < π}. (b) the sector onto the strip Ω 1 = {re iθ : < r <, < θ < π 4 } Ω = {x + iy : < y < 1}. 1
MATH5685 Assignment Supplementary Q1. For each of the following infinite products, determine whether they converge. If they do converge, determine, if possible, what the product converges to. (a) (b) (c) n 1 n + 1. n= ( 1 + in n n + 1 n +. ). Q. Consider the series S(z) = sin(z) n (1 + cos(πz)). (a) Prove that this series does not converge uniformly on C. (b) Prove that the series does converge uniformly on compact subsets of Ω = C \ {±1, ±3, ±5,... }.
Solutions to Assignment 3 Question 1 We showed in class (Corollary ) that if α n converges then (1 + α n ) converges. This obviously implies the if part of the statement. (Or use the hint, as was done in the proof of Theorem 1.) Suppose then that a n and that P = (1 + a n) converges. For N a positive integer let N P N = (1 + a n ). As each term is at least one, the partial product sequence {P N } is increasing and bounded above by the infinite product P. Note that on expanding the finite products P P N = 1 + N a n + a n a m + N a n as all the terms are non-negative. bounded above and hence they converge. Thus the partial sums N a n are increasing and Question Let P N (z) = N ( ) 1 + z n. n= If we again expand out this finite product and reorder the terms it is not hard to see that P N (z) = 1 + z + z + + z N+1 1. (If this is hard for you to see, try proving it by induction.) It follows therefore that as N for all z in the open unit disk. P N (z) k= z k = 1 1 z 3
Question 3 First note that as we showed in class, the integral defining f certainly converges whenever Re z >. Indeed, for z, w in the right half-plane, f(z) f(w) = e t (t z 1 t w 1 ) dt = e t t z 1 (1 t w z ) dt. With some effort one can show that this goes to zero as w z and hence that f is continuous. To show that f is analytic we shall use Morera s Theorem. Let γ be a simple closed contour in Re z >. Then f(z) dz = e t t z 1 dt dz Now the function γ = γ b a e t t γ(s) 1 γ (s) dt ds. G(s, t) = e t t γ(s) 1 γ (s) is at least piecewise continuous and hence it is integrable on [a, b] (, ). We can therefore use Fubini s Theoremto reverse the order of integration: γ f(z) dz = = = b e t t γ(s) 1 γ (s) ds dt. a e t t z 1 dz dt γ e t dt = since the inner integral is zero by the analyticity of z t z 1. It now follows from Morera s Theorem that f is analytic on the right half-plane. Question 4 n 1 ( ) k Let p(n) = Γ. A little experimenting with MAPLE should allow you to guess n k=1 that the right formula is p(n) = (π)(n 1)/ n. To prove this we use the formula that for z C \ Z, Γ(z)Γ(1 z) = 4 π sin πz.
Apart from possibly the middle term (if n is even), the terms in the product come in pairs like this: ( ) ( ) k n k π Γ Γ = n n sin(kπ/n). Thus, if n is odd, and m = n/, p(n) = m k=1 π sin(kπ/n). One can now use this to get the desired formula. Question 5 The first thing to note is that f is the square of the Möbius transformation z g(z) = 1 + z. It is easier to break mapping questions about f into questions about g and about the squaring function s(z) = z. (a) Since g is a Möbius transformation we can find the image of D under g by looking at where the circular boundary goes. In this case g(1) = g( 1) =, g(i) = 1 + i,, and so the unit circle maps to the line Re z = 1. As g() =, this means that D maps to the half-plane H = {z : Re z < 1 }. It now remains to find s(h). But the image of H = {z : Re z } is already the whole of the complex plane, and hence f(d) = s(h) = C too. (b) Now we can use the fact that g(i) = g(i) = + i, g() =, g( i) = i to see that the image of the imaginary axis is the circle 1 with centre 1 and radius 1. As g(1) = 1 this means that the image of the right half-plane is the interior of this circle. Now the squaring function s is one-to-one and continuous on the right half plane, so to find the image of this circle under s it is sufficient to look at where the boundary goes. This bit is actually a little trickier. A MAPLE plot makes it look like the image is a cardiod. With a little trial and error one can confirm this. One can parametrize the imaginary axis by i cot(θ/) for π < θ < π. A nontrivial MAPLE calculation then shows that f(i cot(t/)) = (1 + cos θ)e iθ which traces out the cardiod r = 1 + cos θ. 1 A pedant might complain that g( ) = is not in the image. 5
Question 6 (a) Let f 1 (z) = π ( z ) 1. This clearly maps S to the square S1 = {x + iy : π < x < π, < y < π}. Using Figure 11 on page 37 of Brown and Churchill, the image of S 1 under f (z) = sin z is the upper half ellipse E = { x + iy : y >, x cosh π + } y sinh π < 1. The remaining step is to make this half ellipse to a half disk. There is no nice elementary function which does this. You can find the construction of a suitable function in Gabor Szego, Conformal Mapping of the Interior of an Ellipse onto a Circle, Amer. Math. Monthly, 57 (195), 474 478. The function has the formula f 3 (z) = k 1/ sn ( B sin 1 z, k ) for certain constants k and B, where sn is one of Jacobi s ellipic functions. This is defined as follows. Let φ dθ u(φ) = 1 k sin θ. This has an inverse function φ k (u). The Jacobi elliptic function sn is define by sn(u, k) = sin(φ k (u)). In any case, this does the job! The final function is then the composition f 3 f f 1. (b) Let s first reduce to a more standard situation by applying f 1 (z) = z/ to map Ω 1 conformally onto D 1, the first quadrant of D. Let f (z) = z + 1. Over the whole punctured plane this is not 1 1, but it is not hard z to check that f (z) = f (w) z w = 1 w 1 z = z w zw w = z±1. In particular, f is 1 1 and conformal on D 1. The image is the open 4th quadrant Q 4 = {re iθ : π < θ < }. (See Figure 16, p. 374, of B&C). Now let f 3 (z) = Log z. This maps Q 4 conformally onto the strip S = {x + iy : π } < y <. Finally f 4 (z) = z/π maps S onto Ω. Therefore a suitable conformal map from Ω 1 to Ω is f = f 4 f 3 f f 1. 6
Summary Basic Revision The Principle of the Argument, Rouché s Theorem, Maximum Modulus Theorem, Morera s Theorem. Defining functions via integrals: part 1 (Motivation) Analytic continuation Analytic continuation theorem (Theorem 1), consequences, what it doesn t say! Uniform convergence Main theorem: Theorem 1 about limits of sequence of analytic functions in the topology of uniform convergence on compact sets, Weierstrass M-test. Infinite products The definition, Convergence Tests (Theorems 17 ), the product sin πz for, examples and counterexamples. Products of functions, Theorem 1, πz Corollary 1 (Logarithmic derivative of the product). Functions with a specified sequence of zeros, Weierstrass product theorems (Theorems 4 and 5), Weierstrass Factorization Theorem (Theorem 7). Meromorphic functions, Theorem 8. Γ(z) and ζ(z) Our definition of Γ(z) as a product. Gauss s formula, Functional equations. Euler s formula. The Open Mapping Principle and the Inverse Function Theorem OMP (Theorem 37), Local invertibility if f (z ), Examples of where f (z) on a region but f is not invertible, conformal mappings, IFT (Theorem 41), The Riemann sphere, and analyticity on C. Möbius Transformations Properties, Automorphisms of C and C, cross-rations and their use. The Riemann Mapping Theorem Schwarz s Lemma (Theorem 5), conformal selfmaps of the disk (Theorem 53), what some of the elementary maps to. Riemann Surfaces Complex manifolds, being holomorphi, Riemann surfaces for functions. Elliptic functions What the set of periods can look like (Theorem 6), poles and zeros of elliptic functions (Lemma 64), that they form a field. Construction of elliptic functions Definition of. The type of ODE it satisfies. That and generate all the elliptic functions. 7