Computation of $\int_{0}^{2\pi}\frac{Re^{it}}{z_0+Re^{it}}dt$ without Cauchy Theorem For $z_0\in{\Bbb C}$ such that $z_0\neq0$ and $R<|z_0|$, how can I compute
$$
\int_{0}^{2\pi}\frac{Re^{it}}{z_0+Re^{it}}dt?
$$
With help of Cauchy Theorem, one can conclude that 
$$
\int_{0}^{2\pi}\frac{iRe^{it}}{z_0+Re^{it}}dt=\int_{\gamma}\frac{1}{z}dz=0
$$
where $\gamma=\{z\in{\Bbb C}:|z-z_0|=R\}$.
How can I do this without Cauchy Theorem?

Here is another way: Since $R<|z_0|$, we can write
\begin{eqnarray}
\frac{Re^{it}}{z_0+Re^{it}} &=& \frac{R}{z_0}\frac{e^{it}}{1+\frac{R}{z_0}e^{it}}  \\
&=&  (\frac{R}{z_0}e^{it}) \sum_{n=0}^\infty (-1)^n (\frac{R}{z_0}e^{it})^n \\
&=& \sum_{n=1}^\infty (-1)^{n+1} (\frac{R}{z_0}e^{it})^n \\
&=& \sum_{n=1}^\infty (-1)^{n+1} (\frac{R}{z_0})^ne^{int}
\end{eqnarray}
Since the convergence is absolute, we have $\int_0^{2 \pi} \frac{Re^{it}}{z_0+Re^{it}} dt = \sum_{n=1}^\infty (-1)^{n+1} (\frac{R}{z_0})^n \int_0^{2 \pi} e^{int} dt$, and since $\int_0^{2 \pi} e^{int} dt = 0$ as long as $n\neq 0$, we have the desired result.
