COMPLEX ANALYSIS HW 3

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1 COMPLEX ANALYSIS HW 3 CLAY SHONKWILER 6 Justify the swap of limit and integral carefully to prove that for each function f : D C continuous on an open set D C, and for each differentiable curve : [0, D, the function g defined on C\([0, ) defined by is holomorphic, with derivatives g(z) : w z dw g (n) (z) n! dw. (w z) n+ Proof. Let ɛ > 0. Let z C\([0, ). Let d be the positive distance from z to the curve and let δ d/. Let l() denote the length of and let M be the maximum of f on ([0, ) (such a maximum exists since the image{ of a compact set, like [0,, is compact). Let C such that min ɛδ 3 Ml() }., δ Then, for any w [(0, ), w (z + ) w z + z (z + ) (w z) ((z + ) z) Now, we want to show that [ g(z + ) g(z) lim 0 w z z + z w z δ δ δ. w z dw 0.

2 CLAY SHONKWILER First, note that g(z + ) g(z) [ [ and so g(z + ) g(z) (w z) dw w (z + ) dw w z dw (w z) (w (z + )) dw (w z)(w (z + )) (w z)(w (z + )) dw (w z)(w (z + )) dw [ (w z)(w (z + )) (w z) dw (w z) (w (z + )) (w z) dw (w (z + )) (w z) (w (z + )) dw Now, since < δ, (w z) (w (z + )) w z w (z + ) δ δ. Hence, (w z) (w (z + )) dw (w z) (w (z + )) dw M δ 3 dw M δ 3 l() ɛδ 3 M Ml() δ 3 l() ɛ. Since our choice of ɛ was arbitrary, we see that lim g(z + ) g(z) 0 (w z) dw 0, so g is C-differentiable at z with g (z) dw. Since our choice of (w z) z C\([0, ) was arbitrary, we see that g is holomorphic on C\([0, ). Now, suppose, for the sake of induction, that if f is continuous on Ω, and g f z w dw then g (k) k! dw (w z) k+

3 COMPLEX ANALYSIS HW 3 3 for all k n. Then, returning to our original g, we want to show that g (n ) (z + ) g (n ) (z) lim 0 n! dw. (w z) n+ Let zc\([0, ). Note that Hence, g (n ) (z + ) g (n ) (z) () (w z) n (w (z + )) (w z) n (w (z + )) w z (w z) n (w (z + )) (w z) n (w (z + )) w (z + ) (w z) n (w (z + )) (w z) n. (n )! (n )! (n )! (w (z + )) n [ (w (z + )) n + (n )! (w z) n (w (z + )) (w z) n (w z) n (w (z + )) Now, if we let h(w) w (z+), then, since z + / ([0, ) for all sufficiently small (i.e. < δ), h is continuous on Ω so we can apply the induction hypothesis to conclude that h(w) (n )! dw g(n ) (w z) n h (z) [ (n )! /(w (z + )) /(w (z + )) lim dw 0 (w (z + )) n (w z) n dw, so the first term in the right side of () above is thus simply h(w) (n )(n )! dw (n )(n )! (w z) n (w z) n (w (z + )) dw. As 0, w (z + ) w z, so this converges to (n )(n )! dw. (w z) n+ Similarly, the second term in (), (n )! (w z) n dw (n )! (w (z + )) dw, (w z) n+

4 4 CLAY SHONKWILER so () reduces to (n )(n )! (w z) n (w (z + )) dw+(n )! for sufficiently small. Thus, g (n) g (n ) (z + ) g (n ) (z) (z) lim n! 0 dw n! (w z) n+ (w z) n+ dw. (w z) n+ Therefore, we conclude, by induction, that, the desired result does indeed hold for all n N. 30 For each c C, for all r, r R, and for each f H[A(c, r, r ), prove that there exist f H[A(c, r, ) and f H[D(c, r ) such that f f f. Also prove that there exist only one pair of such functions f and f with lim z f (z) 0. Proof. First, note that it suffices to show this for the case where c 0, because if c 0, then we can form the function h(z) f(z c) which will be holomorphic on A(0, r, r ). Now, because f H[A(0, r, r ), f has a Laurent expansion c n w n where Now, define and n c n. z ρ zn+ f (w) f (w) n c n w n c n w n. To see why these are holomorphic in the appropriate regions, recall that z w z w z ρ where r < ρ < w < ρ < r. Now, for w < ρ and z ρ, z w z w z z n0 n0 z ρ ( w ) n w n z z n+ n0

5 where the convergence is uniform since w z z ρ COMPLEX ANALYSIS HW 3 5 z w <. Hence, z ρ [ n0 [ n0 z ρ c n w n n0 w n z n+ w n zn+ which is just f, where we can swap the sum and the integral since convergence is uniform and c n z ρ holds because H[A(0, r z n+ z n+, r ). Since our only condition for uniform convergence was that w < ρ, we see that f H[D(0, ρ ). As we let ρ r the result still holds, so f is, in fact, holomorphic on D(0, r ). Similarly, if w > ρ and z ρ, then z w z w w w z ( z ) n z n w w w w n+ n0 n0 where the convergence of the series is uniform since z w <. Hence, z ρ z w [ z n z ρ w n+ n0 [ z n w n z ρ n0 n c n w n which is just f, where we can swap the sum and integral since convergence is uniform and substituting c n for the integral is appropriate as above. Our only condition for uniform convergence was that w > ρ, so we see that f H[A(0, ρ, ). As we let ρ r the result still holds, so f is holomorphic on A(0, r, ). Finally, note that, in A(0, r, r ), z ρ n0 z w c n w n + f (w) + f (w) f (w) f (w). n z ρ c n w n z w

6 6 CLAY SHONKWILER Note that, since c n is independent of w, lim f (w) 0. w Now, suppose there exist g H[A(0, r, ) and g H[D(0, r ) such that f g g and lim z g (z) 0. Let r < ρ < ρ < r and define f (z) g (z) if z ρ z ρ h(z) ρ ρ (f (z) g (z)) + ρ z ρ ρ (f (z) g (z)) if ρ z ρ f (z) g (z) if z ρ Then, for z ρ, h is certainly holomorphic, since it is just the difference of two holomorphic functions. For z ρ, h is also holomorphic, since it is, again, the difference of two holomorphic functions. Finally, for ρ < z < ρ, h is holomorphic since z ρ ρ ρ and ρ z ρ ρ are holomorphic on this region, as are the differences f g and f g. Since the cases agree on the boundaries of these regions, we conclude that h is entire. Now, let ɛ > 0. Since lim z f (z) 0 lim z g (z), there exist N and N such that, if z N, then f (z) < ɛ/ and, if z N, g (z) ɛ/. Let N max{n, N }. Then, D N (0) {z C : z N} is compact and so, since h is holomorphic on this region, h is bounded on this region. Furthermore, if z > N, h(z) f (z) g (z) f (z) + g (z) ɛ/ + ɛ/ ɛ, so h is certainly bounded outside D N (0). Therefore, we conclude that h is an entire function that is bounded, and so, by Liouville s Theorem, h is constant. Since h(z) 0 as z, it must, in fact, be the case that h 0. Therefore, for z rho, f (z) g (z) and for z ρ, f (z) g (z). Therefore, the power series expansions of f and g in D(0, r ) must be equal (since the power series is completely determined by the behavior of the functions at 0), and so f g everywhere they are defined. Thus, for ρ < z < ρ, 0 h(z) z ρ ρ ρ (f (z) g (z)) + ρ z ρ ρ (f (z) g (z)) z ρ ρ ρ (f (z) g (z)), so f g inside the annulus as well. From all this, then, we conclude that the f and f we found above are unique. 4 Deduce Weierstrass theorem that the uniform limit of holomorphic functions is holomorphic, from Morera s theorem. Proof. Let f n H(Ω) be a sequence of holomorphic functions on some open set Ω such that f n f locally uniformly. Now, if z Ω, there exists

7 COMPLEX ANALYSIS HW 3 7 D r (z) Ω since Ω is open, and so D r/ (z) Ω. Now, D r/ (z) is compact and relatively compact in Ω, so f n f uniformly on D r/ (z). Hence, f Dr/ is continuous; specifically, f is continuous at z. Since our choice (z) of z was arbitrary, we conclude that f is continuous on all of Ω. Now, let : [0, Ω be a closed path in Ω. Then, since each f n is holomorphic on Ω, we know, by the Cauchy-Goursat Theorem, that f n 0. Now, since ([0, ) is compact in Ω, f n f uniformly on. Hence, we are justified in swapping the limit and integral in the below: [ f lim f n lim f n 0. n n Since we saw above that f is continuous on Ω, the hypotheses of Morera s Theorem are satisfied, and so we conclude that f is holomorphic on Ω. The trigonometric functions sin z and cos z are defined by 5 sin z eiz e iz cos z eiz + e iz. Find formulae for the real and imaginary parts of sin(x+iy) and cos(x+iy), as explicit functions of (x, y). What are the zeros of sin z and cos z in the complex plane? Answer: Let z x + iy C. Then sin z sin(x + iy) ei(x+iy) e i(x+iy) e y+ix e y ix e y e ix e y e ix e y (cos x + i sin x) e y (cos( x) + i sin( x)) e y (cos x + i sin x) e y (cos x i sin x) sin x ( e y + e y) + cos x ( e y e y) sin x ( e y + e y) + i cos x ( e y e y). Now, note that, since e r > 0 for all r R, e y + e y > 0 for all y R. Hence, if sin(x + iy) 0, then it must be the case that sin x 0, so x nπ for some n Z. Of course, for sin to have a zero, the imaginary part must also be 0; since cos nπ ± for all n Z, this implies that e y e y 0,

8 8 CLAY SHONKWILER or e y e y ; since the real exponential is monotonic, this only occurs when y y, which is to say when y 0. Hence, the zeroes of the sin function are exactly those real numbers of the form nπ for n Z. Turning to cos, we see that cos z cos(x + iy) ei(x+iy) + e i(x+iy) e y+ix + e y ix e y e ix + e y e ix e y (cos x + i sin x) + e y (cos( x) + i sin( x)) e y (cos x + i sin x) + e y (cos x i sin x) cos x ( e y + e y) + i sin x ( e y e y). Again, e y +e y > 0 for all y, so cos z has a zero only when cos x 0; namely, when x (n+)π. Of course, the imaginary part must also be zero; since sin (n+)π ±, this occurs only when e y e y 0, which, as we saw above, only occurs when y 0. Hence, the zeroes of cos z are exactly those real numbers of the form (n+)π for n Z. 6 The function tan z sin z cos z. Find the differential equation satisfied by its inverse functions arctan w. As tan 0 0, there is a unique branch of arctan defined in a neighborhood of 0, which satisfies arctan 0 0. Find the power series for this branch of arctan. Answer: First, note that d sin z d ( e iz e iz ) ( ie iz + ie iz) eiz + e iz cos z and Hence, d tan z d cos z d d ( e iz + e iz ) ( ie iz ie iz) eiz e iz sin z. ( ) sin z cos z cos z cos z + sin z cos z sin z + sin z cos z cos z. Therefore, since tan 0 sin 0 cos 0 0 and d tan z z0 cos 0,

9 COMPLEX ANALYSIS HW 3 9 we know, by the inverse function theorem, that tan has a local inverse, called arctan, in a neighborhood Ω of 0 such that arctan 0 0. Furthermore, for z Ω, tan(arctan z) z, and so d tan(arctan z) cos (arctan z) d arctan z. Note that, for any w C, such that cos w 0 (i.e. where R(w) (n+)π ), tan w + sin w cos w + cos w cos w sin w + cos w cos w Hence, the above reduces to so Now, for z <, (tan (arctan z) + ) d arctan z d arctan z d arctan z + z. + z ( z ) n n0 cos w. (z + ) d arctan z, ( ) n z n, which converges absolutely for z <. Therefore, we can integrate termby-term, so we see that the power series for arctan z on D (0) is ( ) n z n+ arctan z n + n0 7 Let f be an entire function that satisfies f(z + ) and f(z + i) for every z C. Prove that f is a constant. Proof. Note that, if n Z, then f(z+n) f(z+(n )+) f(z+(n )) f(z+(n )+) f(z+(n )) and f(z+in) f(z+i(n )+i) f(z+i(n )) f(z+i(n )+i) f(z+i(n )). Now, let denote the floor function (i.e., for x R, x max{a Z a x}). Then, for any z x + iy, note that x, y Z and so n0 f(x + iy) f( x + (x x ) + i( y + (y y ))) f((x x ) + i y + i(y y )) f((x x ) + i(y y ))

10 0 CLAY SHONKWILER Given how is defined, it must be the case that x x < and y y <. Let K {x + iy C 0 x, 0 y }. Hence, sup z C sup. z K Now, since K is compact and f is entire, f K is a continuous function on a compact set and so has both a maximum and a minimum. Specifically, M for some M 0 and for all z K. However, since we just saw that any maximum values of f must occur in K, this means that M for all z C. Therefore, f is an entire, bounded function and so, by Liouville s Theorem, must be constant. DRL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu