Complex Integral $\int_{C}\frac{1}{\sqrt{z}}dz$ I must compute  $\int_{C}\frac{1}{z}dz$ where he branch of $\sqrt{z}$ satisfies $\sqrt{1}=1$.
Where $C$ is a positively oritented semi-cericle $|z|=1$, $0 \leq Arg(z) \leq \pi$.
I am confused on how to satisfy the branch requirement of $\sqrt{z}$. Usually, without branch requirements, I parametrize the contour and continue from there, but with this added requirement I am stuck. 
Apparently, this represents $\sqrt{z}$ on the Riemann Surface (which I am not entire sure what that is)
Can anyone explain what it means to choose a branch of a function like $\sqrt{z}$? Furthermore, how we satisfy the requirement for the above exercise?
 A: By "branch", we make a choice as to whether by "1", we mean $e^{i 0}$ or $e^{i 2 \pi}$, for example.  In this case, $\sqrt{e^{i 0}} = e^{i 0} = 1$ while $\sqrt{e^{i 2 \pi}} = e^{i \pi} = -1$.  So, by stating that $\sqrt{1}=1$, the problem means that $1=e^{i 0}$.  
Thus, being on a single branch, we may then let $z=e^{i \theta}$, $\theta \in [0,\pi]$, and the integral becomes
$$i \int_0^{\pi} d\theta \, e^{i \theta} \, e^{-i \theta/2} = i \int_0^{\pi} d\theta \, e^{i \theta/2} = 2 \left (e^{i \pi/2} - e^{i 0/2} \right ) = 2 (i-1) $$
NB if we were on the other branch, i.e., $\sqrt{1} = -1 = e^{i \pi}$, then $1=e^{i 2 \pi}$ and we would have as the integral
$$i \int_{2 \pi}^{3 \pi} d\theta \, e^{i \theta/2} = 2 \left (e^{i 3 \pi/2} - e^{i 2 \pi/2} \right ) = 2 (-i +1)$$
A: A branch of $\sqrt z$ is defined whenever you take a slit off the complex plane. In your case, you can consider taking away the slit corresponding to $arg(z)=3\pi/2$. Then your function is defined on a simply connected domain and you can applies the fundamental theorem of calculus. 
\begin{equation}
\int_C \frac{1}{\sqrt z}=-2\sqrt1+2\sqrt{-1}=2(i-1)
\end{equation}
