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?

$-i(\ln(z_0 + Re^{it}))' = \dfrac{Re^{it}}{z_0 + Re^{it}}$
So $-i\ln(z_0 + Re^{it})\Bigl|_0^{2\pi} = 0$ because $e^{i0} = e^{i2\pi}$.
