Question #284383

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.


1
Expert's answer
2022-01-03T17:18:02-0500

Cauchy’s integral formula for derivatives:

f(n)(z0)=n!2πif(z)(zz0)n+1dzf^{(n)}(z_0)=\frac{n!}{2\pi i}\int \frac{f(z)}{(z-z_0)^{n+1}}dz


we have:

f(z)=1f(z)=1


then:

for n = 1:

f(z0)=12πi1zz0dz=1f(z_0)=\frac{1}{2\pi i}\int \frac{1}{z-z_0}dz=1


1(zz0)ndz=2πi\int \frac{1}{(z-z_0)^{n}}dz=2\pi i


for n > 1:

f(n1)(z)=0f^{(n-1)}(z)=0


f(n1)(z0)=n!2πi1(zz0)ndz=0f^{(n-1)}(z_0)=\frac{n!}{2\pi i}\int \frac{1}{(z-z_0)^{n}}dz=0


1(zz0)ndz=0\int \frac{1}{(z-z_0)^{n}}dz=0


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