Question involving the Cauchy-Goursat Theorem

Question

Show $\int_C \frac{dz}{(z-z_0)^n} = 2\pi i$ when $n = 1$ and 0 when $n \gt 1$

I'm not entirely sure how to approach this one. Any help would be much appreciated.

Answer 1

Without loss of generality let $z_0 = 0$ and $C$ the unit circle travelled in clockwise direction. Then $$ \int_C \frac{dz}{z^n} = i\int_0^{2\pi} e^{-in\phi} e^{i\phi} d\phi = \frac{i}{1-n} \left. e^{i(1-n)\phi} \right|_0^{2\pi} = 0 $$ for $n \neq 1$. The case for $n = 1$ follows from Cauchy integral formula.

Answer 2

From Cauchy Theorem you have, if $f(z)$ is analytic and if $z_o$ is inside the contour C then $f^n(z_o)$ = $\frac{n!}{2\pi i}$ $\int_C \frac{f(z)}{(z-z_o)(z-z_o)^n}dz$. In your question $f=1$ Now for n=1 , the above reduces to $$f(z_o) = \frac{1}{2\pi i}\int_C \frac{1}{(z-z_o)}dz$$

Since $f(z_o)$ = $1$ hence $\int_C \frac{1}{(z-z_o)}dz$=$2\pi i$.

For n>1, you will have to differentiate $f(z)$ which will be zero since f is constant here.