Branch Cut Issues I'm trying to evaluate what seems to be a straightforward contour integral:
$$I=\int_{\gamma} \frac{dz}{\alpha + \beta z} $$
where $\gamma (t) = e^{-it}$, $t \in \left[ 0,\pi\right]$, $\alpha, \beta \in \mathbb{C}$, and $|\alpha| \ne |\beta|$. Explicitly,
$$I=\int_{0}^{\pi} \frac{(-ie^{-it})dt}{\alpha + \beta e^{-it}}$$
$$=-i\int_{0}^{\pi} \frac{dt}{\alpha e^{it} + \beta }$$
I want to say that
$$I=\frac{1}{\beta}\left(\log\left(\frac{\beta -\alpha}{\beta +\alpha}\right)-i \pi \right)$$
but I feel like I'm neglecting some subtle branch cut issues. Any help or insight would be greatly appreciated.
 A: You do have to be careful to use a branch of the log that is analytic in a neighbourhood of the contour, which in this case is the semicircle in the lower half plane from $1$ to $-1$.
Let's assume $\beta \ne 0$ (the case $\beta = 0$ being very easy).
The integrand has a pole at $z = -\alpha/\beta$, which is not on the contour since $|\alpha| \ne |\beta|$.  Now $\frac{\log(\alpha + \beta z)}{\beta}$ is an antiderivative, but where to put the branch cut?  The branch cut must be a curve from the branch point $z = -\alpha/\beta$ 
to $\infty$.  If you use the branch of log whose imaginary part is from $\tau$ to $\tau + 2 \pi$, the branch cut has $\alpha + \beta z = r e^{i\tau}$, i.e. $z = (-\alpha + r e^{i\tau})/\beta$, for $0 < r < \infty$.  If $\beta = b e^{i\theta}$ with $b > 0$, that extends out from $-\alpha/\beta$ in the direction of argument $\tau - \theta$.  You have to make sure that doesn't hit the semicircle.  For example, if $-\alpha/\beta$ is in the blue region shown below, $\tau - \theta = \pi/2$ would be safe, while in the red region $\tau - \theta = -\pi/2$ would be safe.
Thus with $\alpha = 1$ and $\beta = -2 i$, $-\alpha/\beta = -i/2$ is in the blue region, $\theta = -\pi/2$, and we can take $\tau = 0$: an antiderivative is $F(z) = -\frac{i}{2} \log(1 - 2 i z)$ where
the imaginary part of log is between $0$ and $2 \pi$.  The integral is then
$$ \eqalign{
F(-1) - F(1) &= \frac{i}{2} \left(\log(1 + 2 i) - \log(1 - 2 i)\right)\cr &=
\frac{i}{2} \left( \frac{\ln 5}{2} + \arctan(2) i - \frac{\ln 5}{2} - (2 \pi - \arctan(2)) i \right)\cr & = \pi -\arctan(2)\cr}$$

