So, the integral $$ \mathcal{I}_\gamma(z) = \dfrac{1}{2\pi i}\int_\gamma \dfrac{1}{w-z}dw $$ gives the winding number of $\gamma$ around $z$. This should coincide with the number of turns that the curve $\gamma$ describes around $z$ (minding orientation...)
If I take the curve $\gamma(t)= e^{it}$, $t\in[-\pi,\pi]$ and $z=\frac{1}{2}$, then since $z$ is in the interior part of $\gamma$ one should get $$ \mathcal{I}_\gamma\left(\frac{1}{2}\right) = 1 $$ but when I try to integrate it I keep getting zero (using principal branch of logarithm: $Log(z) := Log(re^{i\theta}) = ln(r) +i\theta ,\theta\in[-\pi,\pi] )$ $$ \begin{array}{rcl} \mathcal{I}_\gamma\left(\frac{1}{2}\right) &=& \dfrac{1}{2\pi i}\displaystyle\int_\gamma \dfrac{1}{w-z}dw \\ &=& \dfrac{1}{2\pi i}\displaystyle\int_{-\pi}^{\pi} \dfrac{ie^{it}}{e^{it}-\frac{1}{2}}dt\\ &=& \dfrac{1}{2\pi i }Log\left(e^{it} - \frac{1}{2}\right)\displaystyle\vert_{-\pi}^{\pi}\\ &=& \dfrac{1}{2\pi i }Log\left( (-1) - \frac{1}{2} - \left( (-1) - \frac{1}{2} \right)\right)\\ &=& \dfrac{1}{2\pi i }\left( ln(1-\frac{1}{2}) + i(\pi) - ( ln(1-\frac{1}{2}) +i(-\pi)) \right)\\ &=& 1. \end{array} $$
I actually can't believe I got this right. I was getting zero before. I'm changing this question to a solution check.