# Parametrization of a line gamma so that $\int_\gamma \frac{1}{2z}dz$ equals $\pi i$

This is a contour integral from a proof, somebody there told me it followed from cauchy integral formula, i know (i probably dont understand it) cauchys integral formula: $$f(z_0)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-z_0}$$and i dont understand how one can get to this :

$$\int_\gamma \frac{1}{2z}dz =\pi i$$

with the help of it. Please do tell me if you see it.

-
think about f(z) = 1 and write $\int_\gamma \frac{1}{z}dz = 2\pi i$ – Integral Sep 12 '12 at 22:01
for $\gamma$ the unit circle (traversed once, counter-clockwise) we have $$\int_{\gamma}dz/z=\int_0^{2\pi}e^{-i\theta}ie^{i\theta}d\theta=2\pi i$$ – yoyo Sep 12 '12 at 22:26

Take $f = 1$, and $z_0 =0$. Then $$1 = \frac{1}{2\pi i}\int_{\gamma}\frac{1}{z-0} = \frac{1}{\pi i}\int_{\gamma}\frac{1}{2z}$$ Thus, $\pi i = \int_{\gamma}\frac{1}{2z}$