Trick of finding branch points While doing some practice problems, I found that the branch points of some functions are the points where they are not defined and at infinity. For example, for $\sqrt z$, the branch points are 0 and infinity. At 0 the function is undefined. For $log(z-1/z+1)$, the branch points are -1, 1 and infinity. Is this a coincidence? Can I use this trick? Thanks.
 A: For complex functions, a singularity is where a function fails to be analytic. Being analytic at a point means having a derivative and being single valued in a neighborhood around that point. All branch points, by definition, are next to multiple valued points. A function therefore fails to be analytic at a branch point, meaning that all branch points are singularities. The converse is not necessarily true: a singularity does not have to be a branch point. Some examples: $\sqrt{z}$ has branch points at $0$ and $\infty$, though it might seem to be well behaved at zero. And $\log{[(z-1)/(z+1)]}$ (which I think is what you meant, see note) does not have a branch point at $\infty$, only $\pm1$. Also, think about $1/z$. It clearly has a singularity at the origin, yet it has no branch points.
While it might be tempting to think that singularities must always be branch points, it is not true in general. The trick to identifying branch points is to know the branch points of basic functions and then cast complicated functions into those basic forms. For example, when we see $\sqrt{z-1}$, we simply let $w = z-1$ to turn it into $\sqrt{w}$. We know $\sqrt{w}$ has a branch point at zero, and when $w=0$, $z=1$, so there is a branch point at $z=1$. Checking the point at infinity requires a bit more cunning. One way is to make the substitution $x=1/z$ and then see if $x=0$ is a branch point, which would imply that $z=\infty$ is a branch point.
Note: $\log{\left(z-\frac{1}{z}+1\right)}$ is an interesting case and turns out to have branch points at $0$, $(-1\pm\sqrt{5})/2$, and $\infty$. The negative root is the golden ratio. Fun to see it pop up.
