Given an annulus around $0$ with the radii $a$ and $b$, can there be a holomorphic square root of $z$?
The proof given is : No, because assuming it existed
$ g(z)^2 = z $ it would follow that : $ 2g(z)g'(z)=1$ and then
$\int _ \gamma \frac{g'(z)}{g(z)} \, dz = \int_\gamma \frac{1}{2z}dz = \pi i$
contradiction. So there cant be a holomorphic square root of $z$ in that annulus.
How does one get to the third step from the second step? And why is it a contradiction?