Question

3. 20 marks] Compute these integrals, with γ a circle of radius 2, Centre at origin, oriented counterclockwise: 2 2z (z-1)3, 22 42 γ ~2 + 8.

Answer 1

(a) Let f(z) = z^2/z - 1 f(z) has a simple pole at z = 1 which lies inside gamma Res_z = 1 f(z) = Lim_z rightarrow 1 (z - 1)f(z) = Lim_z rightarrow 1 z^2 = 1 By Cauchy residue theorem integral_gamma f(z)dz = 2 pi i Res_z = 1 f(z) integral_gamma z^2/z - 1 dz = 2 pi i (b) Let f(z) = e^2z/(z - 1)^3 = phi(z)/(z - 1)^3 where phi(z) = e^2z because phi(1) = e^2 notequalto 0 and phi(z) is analytic at z = 1 therefore z = 1 is a pole of f(z) of order 3 Res_z = 1 f(z) = phi^''(1)/2! Phi^'(z) = 2e^2z, phi^''(z) = 4e^2z phi^''(1) = 4e^2 Res_z = 1 f(z) = 1/2 middot 4e^2 = 2e^2 \r\n By Cauchy residue theorem integral_gamma f(z)dz = 2 pi i middot Res_z = 1 f(z) therefore integral_gamma e^2z/(z - 1)^3 dz = 4 pi i e^2 (c) Let f(z) = 1/z^2 + 4z = 1/z(z + 4)