Originally Posted by

**Erdos32212** Show that if $\displaystyle \gamma$ is a closed curve in the complex plane, with trace $\displaystyle \Gamma$, and if $\displaystyle z_0 \not \in \Gamma$, then $\displaystyle \int_{\gamma} \frac{1}{(z-z_0)^n}dz=0$, $\displaystyle n \geq 2$.

Hint:

For $\displaystyle n \geq 2, (z-z_0)^{-n}$ has primitive $\displaystyle \frac{(z-z_0)^{-n+1}}{1-n}$ on $\displaystyle D$.

Attempt:

So the hint says that the primitive (the antiderivative) is $\displaystyle \frac{(z-z_0)^{-n+1}}{1-n}$. However, I do not see how this will help with showing that $\displaystyle \int_{\gamma} \frac{1}{(z-z_0)^n}dz=0$, $\displaystyle n \geq 2$. Thanks in advance.