Math 6350 Homework #5 Solutions 1. Ahlfors pg. 108 #2: Compute Z x dz |z|=r for the positive sense of the circle, in two ways: first, by use of an explicit parameterization of the circle, and second, by observing that r2 1 1 z+ . x = (z + z) = 2 2 z Solution: For the first method, use z = reit for t ∈ [0, 2π]. We get x = r cos t and dz = ireit dt, so that Z Z 2π x dz = ir2 cos teit dt |z|=r 0 = ir 2 Z 2π cos2 t + i sin t cos t dt 0 = ir2 (π + 0) = ir2 π. For the second method, we note that Z Z 1 r2 x dz = z+ dz 2 C z C Z dz r2 = 2 C z r2 = (2πi) 2 = ir2 π, R using the fact that z is analytic while C dz/z = 2πi whenever C contains the origin. 2. Ahlfors pg. 108R #7: If P (z) is a polynomial and C denotes the circle |z − a| = R, what is the value of C P (z) dz? Answer: −2πiR2 P 0 (a). Hint: adapt the trick from problem #2. Solution: On the given circle we have (z−a)(z−a) = R2 , so that (z−a)dz+(z−a)dz = 0. Hence we have z−a R2 dz = − dz = − dz. z−a (z − a)2 We therefore see that Z P (z) dz = −R C 2 Z C 1 P (z) dz. (z − a)2 We may express P (z) = P (a) + P 0 (a)(z − a) + N (z)(z − a)2 where N (z) is some lower-order polynomial, and we therefore have Z Z P (a) P 0 (a) 2 + N (z) dz = −2πiR2 P 0 (a) P (z) dz = −R + 2 (z − a) z−a C C since the integral of 1/(z − a)2 and the analytic function N (z) are both zero around the circle. R 3. Evaluate the integral γ f (z) dz where f (z) = z|z|2 and γ is the closed upper unit semicircle traversed counterclockwise (that is, rightward along the real axis from −1 to 1, then around the unit circle back to −1). Solution: Write γ = γ1 + γ2 where γ1 is the real axis segment parametrized by z = t on t ∈ [−1, 1] and γ2 is the semicircle parametrized by z = eit for t ∈ [0, π]. We have Z Z 1 Z Z π t4 1 1 1 t dt = = − = 0, f (z) dz = −1 4 4 4 −1 γ1 while 3 f (z) dz = γ2 Hence coincidentally we have 0 R γ t=π eit (ieit ) dt = 2i e2it t=0 = 0. f (z) dz = 0 although f is not analytic. 4. Prove Green’s Theorem for a rectangle R = [a, b] × [c, d]: Z ZZ ∂p ∂q p dx + q dy = − dx dy. ∂x ∂y ∂R R Do not use Stokes’ theorem: do the calculation directly using the Fundamental Theorem of Calculus. Since we will not be using this result at the moment, you may assume that p and q are as smooth as needed to interchange order of integration. Solution: It’s easier to work with the right side. We have Z b Z d Z d Z b ZZ ∂p ∂p ∂q ∂q − (x, y) dx dy − (x, y) dy dx dx dy = ∂x ∂y c a ∂x a c ∂y R Z d Z b = q(b, y) − q(a, y) dy − p(x, d) − p(x, c) dx. c a R We now check that this is γ1 +γ2 −γ3 −γ4 p dx + q dy, where ∂R = γ1 + γ2 − γ3 − γ4 , for γ1 the bottom side, γ2 the right side, γ3 the top side (traversed negatively), and γ4 the left side (also traversed negatively). For γ1 we have x = t for t ∈ [a, b] and y = c, so that dx = dt and dy = 0; we obtain Z Z b Z b Z b p dx + q dy = p(t, c) dt + q(t, c) · 0 dt = p(t, c) dt. γ1 a a 2 a For γ3 we have x = t for t ∈ [a, b] and y = d, so that dx = dt and dy = 0 again, and Z Z b p(t, d) dt. p dx + q dy = a γ3 The vertical paths RR are similar; we see that the signs work out in exactly the same way as they do in R (qx − py ) dx dy. 3

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