How to calculate this complex integral? A homework:
Calculate the integral 
$$\int_{|z|=r}\frac{|dz|}{z-1}$$, $z$ is a complex variable, and $r\neq1$.
When $r<1$, it can be obtained by the mean value formula, I get stacked in the case $r>1$.
 A: Hint:
Use polar coordinates $z=r{e^{i\varphi}}$ on the circle $|z|=r \ $;  then
$$|dz|=r\ d\varphi, \quad 0 \leqslant \varphi< 2 \pi,\\
\int\limits_{|z|=r}\frac{|dz|}{z-1}=\int\limits_{0}^{2\pi}\frac{r\ d\varphi}{r{e^{i\varphi}}-1}.$$
\begin{gather}
\int\limits_{0}^{2\pi}\frac{r\ d\varphi}{r{e^{i\varphi}}-1}=\int\limits_{0}^{2\pi}\frac{r\ d\varphi}{r{e^{i\varphi}}\left(1 - \frac{e^{-i\varphi}}{r}\right)}.
\end{gather}
In the case $r>1$ we can expand
\begin{gather}
\frac{1}{1 - \frac{e^{-i\varphi}}{r}}=\sum\limits_{k=0}^{\infty}{\frac{e^{-ik\varphi}}{r^k}}.
\end{gather}
Note that series is absolutely convergent, so we can integrate it:
\begin{gather}
\int\limits_{0}^{2\pi}\frac{r\ d\varphi}{r{e^{i\varphi}}\left(1 - \frac{e^{-i\varphi}}{r}\right)}=\sum\limits_{k=0}^{\infty}\int\limits_{0}^{2\pi}{\frac{e^{-i(k+1)\varphi}}{r^k}d\varphi}.
\end{gather}
Since $k \ne -1, \quad \int\limits_{0}^{2\pi}e^{-i(k+1)\varphi}\ d\varphi=0,$ therefore, $$\int\limits_{0}^{2\pi}\frac{r\ d\varphi}{r{e^{i\varphi}}\left(1 - \frac{e^{-i\varphi}}{r}\right)}={0}$$
