COMPLEX ANALYSIS HW 8

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1 OMPLEX ANALYSIS HW 8 LAY SHONKWILE 23 For differentiable (but not necessarily holomorphic functions f, p, q : Ω, prove the following relations: (23.: dz d z 2idx dy d z dz. Proof. On one hand, dz d z ( idy (dx idy dx (dx idy + idy (dx idy On the other hand, dx dx idx dy + idy dy dy 2idx dy. d z dz (dx idy ( idy [dx ( idy idy ( idy] [dx idx dy idy dy dy] 2idx dy. Hence, we see that dz d z 2idx dy d z dz. (23.2: dz dz 0 d z d z. Proof. On one hand, dz dz ( idy ( idy dx ( idy + idy ( idy dx idx dy + idy dx dy dy idx dy idx dy 0. On the other hand, d z d z (dx idy (dx idy dx (dx idy idy (dx idy dx dx idx dy idy dx dy dy idx dy + idx dy 0. Hence, we see that dz dz 0 d z d z. (23.3: ( + f df.

2 2 LAY SHONKWILE Proof. First, note that ( f i f f z fdz 2 Also, f z fd z 2 (dx+idy 2 [ ( ] f f f dy + i f dy dx. ( f + i f (dx idy [ ( ] f f f 2 dy + i f dx dy. Therefore, ( + f [ ( ] f f f 2 dy + i f dy [ ( ] f f f 2 dy + i f dx dy f f dy df. (23.4: 0. Proof. Note that ( f ( z fdz z z fdz dz 0 by our result in (23.2 above. Similarly, ( f ( z fd z z z fd z d z 0, again by our result in (23.2. Hence, since the choice of f is arbitrary, we see that 0. (23.5: /4 dz d z. Proof. ecall that f [ ( ] f f f 2 dy + i f dx dy as we saw in (23.3. Hence, ( f ( z fd z z z fdz d z [ ( f z 2 + i f [ 2 f i 2 f i [ 2 f f 2 fdz d z. 4 ] dz d z ] dz d z ( 2 ] f + f i 2 2 dz d z

3 On the other hand, ( f ( z fdz z z fd z dz [ ( f z 2 i f [ 2 f 4 2 i 2 f + i [ 2 f f 2 OMPLEX ANALYSIS HW 8 3 fd z dz 4 fdz d z. 4 ] d z dz ] d z dz ( 2 ] f f i 2 2 d z dz Hence, we see that, since our choice of f was arbitrary, /4 dz d z. (23.6: ( + [( + f] 0, so that d(df 0 and d d 0 with complex functions also. Proof. ecall, from (23.3, that Hence, ( + [( + f] d(df ( f d ( + f f f dy df. f dy ( 2 f 2 2 f dy ( 2 f 2 f 2 dx 2 f dy 2 f dx dy + 2 f dx dy 0, 2 dy 2 f dy ( 2 f dx dy + 2 f dy dy 2 since 2 f. Therefore, since our choice of f was arbitrary, we see that d d 0 for complex functions as well. 2 f (23.7: d(pdz + qd z ( z q z pdz d z.

4 4 LAY SHONKWILE Proof. Simply computing, we see that ( p p d(pdz + qd z dy ( idy + p p q q dy i dx dy + dy dx i dx dy ( p + i p ( dx dy + q i q dx dy ( p 2i + i p dz d z + ( q 2i i q ( p 2 i p dz d z + ( q 2 i q dz d z ( z p + z qdz d z. (23.8: z f z f. ( q q dy (dx idy dz d z Proof. Suppose f f + if 2 for real-valued f and f 2. Then ( f f z f + i 2 [ ( ] f f Hence 2 2 [ f z f 2 [ f 2 [( f 2 [ f 2 i f 2 + i [ f + f 2 + i + f 2 + f 2 + i f 2 ] i f z f. i f 2 ( f f 2 ( + i f f 2 ( + i f2 f i ] ] ]. ( f + i f 2 ] 232 For every open set Ω with smooth boundary Ω, and every form ω pdz + qd z, prove that Ω ω Ω dω. Note in particular that if

5 ω pdz, then Proof. First, note that OMPLEX ANALYSIS HW 8 5 Ω pdz p z d z dz. Ω ω pdz + qd z p( idy + q(dx idy (p + q i(p qdy. If is a rectangle then, by Exercise 229, ω (p + q i(p qdy (p + qdy i (p qdx dy [( p + i p ( + q i q ] dx dy [( p ( q 2 i p + i q ] dz d z ( z p + z q dz d z d(pdz + qd z by (23.7 above. Let L ɛ be the lattice {mɛ + inɛ m, n Z}, let Ŝɛ be the set of squares of side ɛ with vertices in L ɛ and let S ɛ { Ŝɛ Ω}. Then dω dω ω S ɛ S ɛ S ɛ by the result proved above, where is traversed counterclockwise for each S ɛ. Now, each edge of a square in S ɛ interior to S ɛ is traversed once in each direction in this sum, so dω ω ω. S ɛ S S ɛ ɛ dω Now, as ɛ 0, and so we see that, in fact, dω S ɛ ω S ɛ Ω Ω Ω Ω ω, dω ω dω.

6 6 LAY SHONKWILE Now, in the specific case mentioned above, when ω pdz, dω d(pdz dp dz ( z + z p dz so, using the result proved above, pdz ω Ω Ω ( z pdz + z pd z dz p z dz dz + p z d z dz p z d z dz, Ω dω p z d z dz. Ω 239 For each positive integer n and each real λ >, prove that the equation z n e z λ has no solutions with z and exactly n simple solutions with z <. Proof. When z, note that z n z n n. On the other hand, if λ >, z and z x + iy, e z λ e x+iy λ e x λ e iy e x λ < since x and λ > and so x λ < 0. Therefore, we see that e z λ < z n when z, so there can be no solutions to the equation z n e z λ with z. On the other hand, if we let Ω, then z n and e z λ are both holomorphic on all of Ω. If we let K D (0 be the closed disc of radius centered at the origin, then we ve just demonstrated that e z λ e z λ < z n on the boundary Ω so, by ouché s Theorem, ord a (z n e z λ. a K ord a z n a K Now, the only zero of z n is at the origin, and the order of z n at the origin is n, so we see that ( n a K ord a (z n e z λ. Hence, z n e z λ 0 must have at least one solution in K; since we ve seen there are no such solutions with z, we know that any solutions must have modulus strictly less than.

7 OMPLEX ANALYSIS HW 8 7 Now, if n, the above implies that z n e z λ has a single simple solution. If n >, z 0 is a solution of z n e z λ and we denote f(z z n, g(z e z λ, then f (z 0 g (z 0 nz n 0 e z 0 λ. Now, nz n 0 n z 0 n > n z 0 n z 0 n z n 0 ez 0 λ, so we see that f (z 0 g (z 0 nz n 0 e z 0 λ 0, meaning that z 0 is a simple solution of z n e z λ. Since our choice of solution z 0 was arbitrary, we see that all solutions of this equation are simple. Therefore, by ( and the remark following it, z n e z λ has n simple solutions with z <. 24 For each non-empty open set Ω and for each continuously differentiable function φ : Ω such that the partial differential equation z u φ in Ω has at least one continuously differentiable solution u : U, identify all the continuously differentiable solutions v of z v φ in Ω in terms of the solution u and other types of functions. Proof. Suppose v : V is a continuously differentiable solution to z v φ. Let h : U V be given by h u v. Then, for z Ω, z h(z, z z (u v(z, z φ(z, z φ(z, z 0, which is to say that h Ω H(Ω. On the other hand, if h : W is such that h Ω H(Ω, then z (u + h z u φ on Ω, so u + h is a solution to the equation z v φ. Hence, we conclude that v is a solution to z v φ if and only if for h H(Ω. Suppose that φ c v Ω u Ω + h 5 ( and let u(z π Fix > 0 such that supp φ. φ(ζdξdη. (a: Show that in the domain z > we have the uniformly convergent expansion a j u(z z j. j0

8 8 LAY SHONKWILE (2 u(z π Proof. Since supp φ, we can reduce to φ(ζdξdη π φ(ζdξdη φ(ζ 2πi d ζ dζ by our work in (23.. Now, since ζ < for ζ, 0 for all z >, so ζ z is a holomorphic function in terms of ζ on ( for any such z; that is, ζ ζ z 0. Hence, ζ ( ( u(ζ ζu(ζ ζ u(ζ + u(ζ ζ. Applying this to (2, we see that (3 u(z ( u(ζ 2πi ζ d ζ dζ 2πi ζ u(ζ dζ by our result proved in Exercise 232 above. Now, z( ζ z z ζ ( ζ n ζ n z z z n+ z n0 n0 since ζ z < and this series converges uniformly. Applying this to (3, we see that u(z u(ζζ n [ 2πi ζ z n+ dζ ] u(ζζ n dζ 2πi n0 n0 ζ z n+, where we can swap the sum and integral since the convergence of the sum is uniform. Hence, where u(z which is of the desired form. n a j z j a j u(ζζ j dζ, 2πi ζ (b: Find formulæ for the coefficients {a j }. Answer: In part (a above we found one formula for a j ; since ζ j is entire for z, we know that ζ(ζ j 0. Hence, ζ ( u(ζζ j ζu(ζ ζ j + u(ζ ζ(ζ j φ(ζζ j,

9 and so, by Exercise 232, a j u(ζζ j dζ 2πi ζ 2πi 2πi π by (23.. OMPLEX ANALYSIS HW 8 9 ζ(u(ζζ j d ζ dζ φ(ζζ j (2idξ dη φ(ζζ j dξ dη, (c: Show that if the equation z f φ has a compactly supported solution, then it is unique. Proof. Suppose v is a solution of z f φ with compact support. Now, by the result proved in Exercise 24 above, we know that any solution f of z f φ is of the form f v + h for some h H(. Now, since v has compact support and no holomorphic function has compact support except 0 (by Liouville s Theorem, this implies that f does not have compact support unless h 0, which means that f v. Hence, v, if it exists, is the unique solution of z f φ with compact support. (d: Find a necessary and sufficient condition for the existence of a compactly supported solution in terms of the {a j }. Answer: Suppose v is a solution to z f φ with compact support. Then v(z π ζv dξdη π Now, by the results found in (a and (b above, a j (4 v(z z j n φ(ζdξdη. for z >. Let Ω {z : z > }. Then, since 0 / Ω, each a j z j is holomorphic on Ω, and so v H(Ω. Since v has compact support, there exists N such that v(z 0 for z > N, which means v 0 on the open subset Ω {z : z > N} Ω. In turn, since v H(Ω, this implies that v 0 on all of Ω. Now, (4 is just a Laurent expansion of v on Ω; by the uniqueness of the Laurent expansion, this implies that a j 0 for all j N. On the other hand, if all the a j are zero, then v 0 on all of Ω, and so v certainly has compact support; specifically, supp v.

10 0 LAY SHONKWILE Therefore, we conclude that the a j s being zero is a necessary and sufficient condition for v to have compact support. (e: When it exists, give a formula for the compactly supported solution. Answer: If v is compactly supported, then v(z π ζv dξdη. Hence, if v is a compactly supported solution to z v φ, then v(z φ(ζdξdη. π 245 For each φ c ( with φ 0 on, prove that if there exists u c ( with u/ z φ, then φ 0. Proof. By 5(e above, if u c ( such that z u φ, then u(z φζdξdη π. Hence, by our results in 5(a and 5(b, if supp φ, then for z >, where a j π u(z j a j z j φ(ζζ j dξ dη. Now, since u has compact support, a j 0 for all j, 2,.... Specifically, 0 a φ(ζdξ dη. π Since φ 0 on all of and, hence, on all of, φ(ζdξ dη 0 φ D (0 0. Since contains the support of φ, this in turn implies that u has compact support only if φ 0.

11 OMPLEX ANALYSIS HW onstruct a connected open set Ω and a function φ c (Ω such that the equation u/ z φ has a solution u c ( but no solution u c (Ω. onstruction: Let f be a function on such that f(z for z and f(z 0 for z 2. For example, z f(z e z z 2 z z < z < 2 0 z 2 is such a function. Then, since f is, so is z f. Let φ z f. Now, since f is constant on {z : z }, φ(z 0 for z. Similarly, since f is constant on {z : z 2}, φ(z 0 for z 2. Therefore, supp φ {z : z 2}, so z f c (. Furthermore, if Ω {z : /2 < z < 5/2} then Ω is a connected open set and φ c (Ω. Now, f is certainly a solution of z f φ and f c (. Suppose there exists u c (Ω such that z u φ. Then, since 0 φ(z z u(z for z {z : z or z 2} Ω, u must be constant on this region; since u is compactly supported on Ω, u must, in fact, be 0 in this region. Thus, we can smoothly extend u to a compactly supported function on all of by letting u(z 0 for z \Ω. Now, if we denote this extended function by û, then z û φ. Now, û(0 0 f(0, so û f. However, since we know f is a compactly supported solution to z v φ, this implies that there are two distinct compactly supported solutions of z v φ, contradicting our result in 5(d above. From this contradiction, we conclude that there is no such u c (Ω and, hence, φ and Ω have the desired properties. DL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu