Branch points and natural maximal domain of $\log\frac{1+z}{1-z}$ So I have a function $\log\frac{1+z}{1-z}$ and I'm supposed to find its branch points, natural maximal domain. 
So far I converted $z$ to $x+iy$ but no real good. Can't check holomorphicity using this. I know that $\log$ is holomorphic after a branch cut. How do I proceed?
I also conclude that $\frac{1+z}{1-z}$ is holomorphic as it doesn't contain $\bar{z}$
 A: We can't allow $0$ or $\infty$ in $\log$, so the points $z=\pm 1$ are out. 
Also: the complex logarithm is multi-valued, acquiring a multiple of $2\pi i$ as its argument moves along a closed curve separating $0$ from $\infty$.  If you don't want this to happen, you must make sure that your domain has no closed curves separating $1$ from $-1$. 
As Daniel Fischer noted, there is no canonical way of making this happen. Cutting out the line segment from $-1$ to $1$ is one natural choice; there is another one with cuts along the real axis (I leave it for you to find it), and plenty of others, which are less nice.
A: We can define
$$
\log\left(\frac{1+z}{1-z}\right)=\frac\pi2i+\int_i^z\left(\frac1{w+1}-\frac1{w-1}\right)\,\mathrm{d}w\tag{1}
$$
Since the residues at $+1$ and $-1$ cancel, the integral along any closed path that circles them both an equal number of times will be $0$.  The smallest branch cut which forces any closed path to circle both an equal number of times is the segment $[-1,+1]$. Thus the largest natural domain would be $\mathbb{C}\setminus[-1,+1]$.
