Let $\gamma(t)=1+e^{it}$. Find $\int_\gamma \big(\frac{z}{z-1}\big)^n\,dz$.

Question

Let $\gamma(t)=1+e^{it}$. Find $\int_\gamma \big(\frac{z}{z-1}\big)^n\,dz$.

So by substituting, I end up with

$$\int_0^{2\pi} i(e^{-it}+1)^ne^{it} \, dt.$$

However, using Cauchy's integral formula, I get just the $(n-1)$th derivative of $z^n$ which is $n!z$ so is the final answer just $2\pi i n$?

Answer 1

By Cauchy's integral formula, if $\gamma(t):=1+e^{it}$, then

\begin{align} \int_\gamma\bigg( \frac{z}{z-1}\bigg)^ndz&=\frac{2\pi i}{(n-1)!}\bigg(\frac{d^{n-1}}{dz^{n-1}} z^n\bigg)\bigg\vert_{z=1}\\ &=\frac{2\pi i}{(n-1)!}n!z\bigg\vert_{z=1}\\ &=2\pi i n. \end{align}