Answer to Question #284383 in Complex Analysis for tayyaba

Question

Suppose z0 is any constant complex number interior to any simple closed

contour C. Show that for a positive integer n,

∮

dz

(z−z0)

n = {

2πi, n = 1

C 0, n > 1.

Answer 1

Cauchy’s integral formula for derivatives:

$f^{(n)}(z_0)=\frac{n!}{2\pi i}\int \frac{f(z)}{(z-z_0)^{n+1}}dz$

we have:

$f(z)=1$

then:

for n = 1:

$f(z_0)=\frac{1}{2\pi i}\int \frac{1}{z-z_0}dz=1$

$\int \frac{1}{(z-z_0)^{n}}dz=2\pi i$

for n > 1:

$f^{(n-1)}(z)=0$

$f^{(n-1)}(z_0)=\frac{n!}{2\pi i}\int \frac{1}{(z-z_0)^{n}}dz=0$

$\int \frac{1}{(z-z_0)^{n}}dz=0$