Key to Homework 8, Thanks to Da Zheng for the text-file

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1 Key to Homework 8, Thanks to Da Zheng for the text-file November 8, 20. Prove that Proof. csc z = + z + 2z ( ) n z 2 n 2, z 0, ±π, ±2π, π2 n= We consider the following auxiliary function, where z 0, ±π, ±2π, f(ζ) = csc ζ ζ ζ(ζ z) Then, it is easy to check that csc ζ ζ has a removable singularity at ζ = 0, so we will regard it holomorphic at ζ = 0. Thus, the function f has poles at ζ = nπ, n Z, each of order one.here, notice that ζ = 0 is also a removable singularity for f. Next, choosing the following sequence of counterclockwise square contour γ n, centered at the origin with vertices (n + 2 )(± ± i)π. Hence, integrating f(ζ) over contour γ n and applying the residue s theorem, we obtain 2πi γn f(ζ)dζ = csc z z z + n ( ) k n kπ(kπ z) + ( ) k kπ(kπ + z) k= Then, for enough large n(i.e. γ n contains z), we evaluate the integral on the left hand side as follows: we first claim that csc z is bounded on the boundary of the square γ n (please see this argument carefully ): Indeed, let z = x + iy, for y > π 2, csc z = 2i e iz e iz 2 e y e y 2. e π/2 e π/2 Similarly, for y < π 2, csc z. In other words, csc z on two horizontal lines of the square and also on the part of two vertical lines with y > π 2. Hence, k= ( )

2 it only remains to see the points on the part of two vertical lines with y π 2. Notice that on the line segment with ends π 2 π 2 i, π 2 + π 2 i, the function sin z has no zeros, so csc z M on the line segment. Since csc(z + π) = csc z, so csc z M on the line segment with ends n + π 2 π 2 i, n + π 2 + π 2 i. Hence csc z M on γ n. Thus f(ζ)dζ = csc ζ γ n γn ζ ζ(ζ z) dζ csc ζ ζ γ n ζ(ζ z) dζ csc ζ γ n ζ(ζ z) + ζ ζ(ζ z) dζ 2πM (n + /2)π z + 2π (n + /2)π((n + /2)π z ) 0, as n + So, the expression (*) becomes 0 = csc z z + ( ) k z kπ(kπ z) + ( ) k kπ(kπ + z) k= k= Simply the above and we obtain our desired equality csc z = + z + 2z ( ) n z 2 n 2 π 2, n= z 0, ±π, ±2π, 2. Find the expression of where a > 0 is not a integer. Solution. + We set the following auxiliary function f(z) = n 2 + a 2 z 2 + a 2 cot(πz) Notice that f has poles at z = ±ai, z = k, k Z, each of order one. Choose a sequence of counterclockwise square contour γ n, centered at the origin with vertices (n + 2 )(± ± i. We claim that cot πz is bounded on γ n. The argument is similar to the above. Indeed, for Rez = y > 0, cot πz = e iπz + e iπz e iπz e iπz e2πy + e 2πy. 2

3 So when y 2, cot πz eπ + e π. Since cot( πz) = cot(πz), when y 2, we also have cot πz eπ + e π. So cot πz eπ + e π on two horizontal lines of the square and also on the part of two vertical lines with y > 2. Hence, it only remains to see the points on the part of two vertical lines with y 2. Notice that on the line segment with ends 2 i 2 and 2 + i 2, cot πz is continuous, so so cot πz M on the line segment. Since cot π(z + ) = csc πz, so csc πz M on the line segment with ends n + 2 i 2 and n i 2. Thus cot πz is bounded on γ n, i.e. cot πz M on γ n. Then, apply the residue s theorem to γ n, where n is large enough, and we get cot ai f(z)dz = + n 2πi γ n ai πa 2 + k= π(k 2 + a 2 ) + k= n π(k 2 + a 2 ) For the integral on the left hand side, and large enough n, by the claim, f(z)dz = γ n γ n z 2 + a 2 cot(πz)dz z 2 + a 2 cot(πz) dz So, we let n +, (*) becomes 0 = cot ai ai This result implies that γ n 2π(n + /2)M (n + /2) 2 a 2 0, as n πa 2 + π(k 2 + a 2 ) + k= k= π(k 2 + a 2 ) ( ) + n 2 + a 2 = π ea + e a a(e a e a ) 3. Page 49, #30 3

4 Calculate the stereographic projection explicitly in terms of the coordinates of three-dimensional space. Solution. Set z = x+yi, and it corresponding stereographic projection: Z = (x, x 2, x 3 ). So the line segment between the north pole: N = (0, 0, ) and z can be written as (( t)x, ( t)y, t), t [0, ] Then, there is a t 0 [0, ], such that Z = ( t 0 )z + tn, i.e. Next, since Z S, we have x = ( t 0 )x, x 2 = ( t 0 )y, x 3 = t 0 ( ) x 2 + x x 2 3 = ( t 0 ) 2 x 2 + ( t 0 ) 2 y 2 + t 2 0 t 2 0 = ( t 0 ) 2 z 2 So, t 0 = or t 0 = z 2 z 2 +. If t 0 =, then Z = N; if t 0 = z 2 z 2 +, then plug t 0 in (*), we have x = ( t 0 )x = z + z z 2 +, x 2 = ( t 0 )y = z z i( z 2 + ), 4. Page 49, #32 x 3 = t 0 = z 2 z 2 +. If z, w are points in the complex plane, then the distance between their stereographic projections is given by d(z, w) = 2 z w ( + z 2 )( + w 2 ) Prove this. The resulting metric on C is called the spherical metric. Proof. Set stereographic projections of z and w to be Z = (x, x 2, x 3 ), W = (y, y 2, y 3 ) Notice that Z, W S, so x 2 + x x 2 3 = y 2 + y y 2 3 =. Next, d(z, W ) = (x y ) 2 + (x 2 y 2 ) 2 + (x 3 y 3 ) 2 d 2 (Z, W ) = 2 2(x y + x 2 y 2 + x 3 y 3 ) 4

5 Combine the results from the above problem, we obtain that ( z + z d 2 (Z, W ) = 2 2 z 2 + w + w w z z ) i( z 2 + ) w w i( w 2 + ) + z 2 z 2 + w 2 w 2 + = zw + 2 wz + z 2 w 2 z 2 w 2 + ( z 2 + )( w 2 + ) = 2 2 z w 2 2 zw 2 wz ( z 2 + )( w 2 + ) = 4 z w 2 ( z 2 + )( w 2 + ) This implies the result d(z, w) = 2 z w ( + z 2 )( + w 2 ) 5. Determine which of the following family is normal. (a). F = {z n } + n= on the unit disc D(0, ). (b). F = { z } + n /2 n= on C. (c). F = {f(z) : f(z) c for some c C} on C. Proof. (a). Since on each compact subset K D(0, ), we have that K D(0, r) with r <, so z r <, then z n r, for each n N i.e. f(z) M K for all z K, f F. Hence, by the Montel s theorem, we know that F = {z n } + n= is a normal family on the unit disc D(0, ). (b). Similarly, on each compact subset K C, we know that z M K (the bound depends on K), then z n /2 M K, for each n N Hence, by the Montel s theorem, we know that F = { z } + n /2 n= on C. (c). F = {f(z) : f(z) c for some c C} on C is not a normal family C, because, if we take the sequence {f n (z) n} n=, then it does not have any convergent subsequence on any compact subset K C. 5

6 6. Let U C be a connected open set. Show that the family is normal. F = {f(z) : f(z)is holomorphic on U and R(f) > 0} Proof.Proof. Since the ball D( 2, ) is disjoint with the right half plane {z Rez > 0}. The function 2+f(z) is holomorphic on U and 2 + f(z) for all f F. Hence, by Montel theorem, the new family F := { 2 + f(z), f F} is normal on U. We now prove that F is normal on U in the general sense. For any subsequence {f n } F. Since F is normal, there exists a subsequence {f nk } of {f n } such that 2+f nk converges uniformly on every compact subset K U to a holomorphic function g. Since 2+f nk has no zeros on U, by Hurwitz s theorem, either g has no zeros on U or g 0 on U. Case. g has no zeros on U. We claim that {f nk } converges uniformly on every compact subset K U to g 2. Fix a compact subset K U. Then c g(z) c 2 on K for some constants c, c 2 > 0. Since 2+f nk converges uniformly on K to g, taking ɛ = c /2, there exists N > 0 such that for all n k > N and z K, g 2 + f nk c /2. Hence or c 2 + f nk g 2 + f nk c /2 + f nk 2 c. ( ) Now, again since 2+f nk converges uniformly on K to g, for every ɛ > 0, write ĝ = /g, there exists N 2 > N such that for all n k > N 2, on K, 2 + f nk ĝ < ɛ, or f nk (ĝ 2) (2 + f nk ) ĝ < ɛ. 6

7 Using (*) and ĝ c 2, we get This proves the claim. f nk (ĝ 2) < 2 c c 2 ɛ. Case 2: g 0 on U. It is easy to show, use the similar argument that {f nk } diverges uniformly on every compact subset K U to. Hence F is normal on U in the general sense. This finishes the proof. Proof 2. Using the following linear fractional transform G(z) = z z + Then, we know that G is a conformal mapping from the right half plane onto the open unit disc D(0, ). Next, consider the following family of holomorphic functions Then, we know that on U, M = {G f(z) : f F} G f(z), for each f F So, by the Montel s theorem, it follows that M is normal family. Again, to simplify the proof, we will focus on the following exhaustive sequence of compact subsets for Ω, named, K K 2 K n. Without loss of generality, we suppose K contains a open disc D(z, δ). Next, to see that F is also a normal family (with generalized definition), we take a sequence of functions of F, say {G f n }, and there is a subsequence, say {G f nk }, and by the above result {G f n (z)} n= is uniformly Cauchy sequence on every compact subset of U. Now, by the Hurwitz theorem, we shall divide the following argument into two cases, by noticing that {G f nk } converges uniformly to g is equivalent to that converges uniformly to g. 2 f nk + Case. g on Ω, then on K n, 2 f nk (z) + < ɛ This directly implies that f nk diverges uniformly on each K n, so f nk diverges uniformly on every compact subset of Ω. 7

8 Case 2. g is nowhere on Ω, then on each K n, there is a c > 0 and c 2 > 0 such that c g c 2. So for c /2, there exists an N N, such that for all n k N, f nk (z) f n (z) + g < ɛ i.e. c 2 f nk + 2 g f nk + c /2 Thus, f nk + 4/c, for n k N. Next, for n k > n p > N, by the uniform Cauchy convergence, 2(f nk f np ) (f nk + )(f np + ) < ɛ But also, 2(fnk f np ) 6 < c 2 ɛ This shows that {f nk } is uniformly Cauchy sequence on K n, thus uniformly converges to a holomorphic function. So, we conclude that f nk converges normally on Ω, which completes the proof. 8