Clarification on choice of branch cut on complex log I am trying to evaluate the following integral, and I have the answers:
$$\int_C \left(e^z + \frac1z\right)\,\mathrm dz$$
Where $C$ is the lower half of the circle with radius $1$ centre $0$, negatively oriented.
So a primitive is: $e^z + \log z$ and now we must choose a branch cut for the complex log, and they choose:
$$\frac\pi2 \lt \theta \lt \frac{5\pi}{2}$$
I will admit, I don't totally understand the branch cut, beyond the idea that it limits us to a $2\pi$ range, and that it gives us a unique solution. Why not just stick with $-\pi \lt \theta \leq \pi$. Actually that raises a second question, why do they set it to $\lt$ both times, this gives me some insight perhaps: Did they choose this, since there is a line they are excluding? Maybe I just worked it out, they are cutting out the positive imaginary axis for $(0,y)$?
 A: You have worked it out.
To your first question, remember that 
$$
\text{log}(z)=\text{log}|z|+i \text{arg}(z)
$$
Then you just need to see that, for a branch of the $\text{log}$ to be a single valued function, it must have its argument in a $2\pi$ length interval. 
For the branch cut, since the path $C$ does not crosses the positive imaginary axis you will want to pick a branch whose domain is 
$$
\mathbb{C} \setminus \{ iy : y \geq 0\},
$$
hence the argument of $\text{log}(z)$ must be in $(\pi/2, 5\pi/2)$ or adding a $\pm 2\pi$ shift. Now note that at the begining of the path $C$  ($z=1$), $\text{log}(z)$ will have an argument of $2\pi$ but at the end ($z=-1$) the argument is $\pi$. Thus, parametrizing $C$ as $C(\theta)=e^{i \theta}$, where $\theta$ runs from $2\pi$ to $\pi$, the integral is 
\begin{align}
\int_C \left( e^z + \frac{1}{z}\right)&=e^{C(\pi)}+\text{log}(C(\pi))-e^{C(2\pi)}-\text{log}(C(2\pi))\\
& = e^{-1}+\text{log}(-1)-e^1-\text{log(1)}\\
& = \frac{1-e^2}{e}+\text{log}|1|+i\pi -\text{log}|1|-i2\pi \\
& =  \frac{1-e^2}{e}-i\pi
\end{align}
