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.
Cauchy’s integral formula for derivatives:
f(n)(z0)=n!2πi∫f(z)(z−z0)n+1dzf^{(n)}(z_0)=\frac{n!}{2\pi i}\int \frac{f(z)}{(z-z_0)^{n+1}}dzf(n)(z0)=2πin!∫(z−z0)n+1f(z)dz
we have:
f(z)=1f(z)=1f(z)=1
then:
for n = 1:
f(z0)=12πi∫1z−z0dz=1f(z_0)=\frac{1}{2\pi i}\int \frac{1}{z-z_0}dz=1f(z0)=2πi1∫z−z01dz=1
∫1(z−z0)ndz=2πi\int \frac{1}{(z-z_0)^{n}}dz=2\pi i∫(z−z0)n1dz=2πi
for n > 1:
f(n−1)(z)=0f^{(n-1)}(z)=0f(n−1)(z)=0
f(n−1)(z0)=n!2πi∫1(z−z0)ndz=0f^{(n-1)}(z_0)=\frac{n!}{2\pi i}\int \frac{1}{(z-z_0)^{n}}dz=0f(n−1)(z0)=2πin!∫(z−z0)n1dz=0
∫1(z−z0)ndz=0\int \frac{1}{(z-z_0)^{n}}dz=0∫(z−z0)n1dz=0
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