calculation of contour integral

We will determine the important complex integral

$$I:={\oint }_C{(z-z_0)}^n𝑑z$$

where $C$ is the circumference of the circle  $|z-z_0|=\varrho$  taken anticlockwise and $n$ an arbitrary integer.

Let’s take the “direction angle” of the radius of $C$ as the parametre $t$, i.e.

$$t:=\arg |z-z_0|.$$

Then on $C$, we have

$$z-z_0=\varrho e^{it},0\leqq t\leqq 2\pi$$

and

$$dz=i\varrho e^{it}dt,{(z-z_0)}^n={\varrho }^n{e}^{int},$$

whence

$$I={\int }_0^{2\pi }{\varrho }^n{e}^{int}i\varrho e^{it}𝑑t=i{\varrho }^{n+1}{\int }_0^{2\pi }{e}^{i(n+1)t}𝑑t.$$

In the case  $n=-1$  one gets trivially  $I=2i\pi$.  If  $n\ne -1$,  we obtain

$$I=i{\varrho }^{n+1}\underset{t=0}{\overset{2\pi }{/}}\frac{e^{i(n+1)t}}{i(n+1)}=\frac{{\varrho }^{n+1}}{n+1}(1-1)=\mathrm{\hspace{0.33em}0},$$

using the fact that $2i\pi$ is a period of the exponential function (http://planetmath.org/PeriodicityOfExponentialFunction).

Hence we can write the result

${\displaystyle {\oint }_C}{(z-z_0)}^n𝑑z=\{\begin{array}{cc}2i\pi \hspace{1em}\text{if}n=-1,\hfill & \\ 0\hspace{1em}\hspace{1em}\text{if}n\in \mathbb{Z}\setminus \{-1\}.\hfill & \end{array}$