How to show that a point is a branch If a function $f(z)$ has a branch point around $\alpha$ than the endpoints of a path around $\alpha$ do not map to the same point under $f(z)$. So we must test for inequality of $f(\alpha + re^{i2\pi}) \neq f(\alpha + re^{0})$. But this method only works for simple functions. For example given a fucntion $$g(z) = \sqrt{1 + \sqrt{z}}$$ Zero is obviously a branch poin, but 1 is also a branch point for (2 out of 4) branches of this function where the inner square root is the branch for which $\sqrt{1} = -1$. I understand why this is, but I don't understand how to algebraically prove/show it?

Rewrite $w=\sqrt{1+\sqrt{z}}$ as $z=(w^2-1)^2$, and note where the derivative of the latter wrt $w$ is zero. You get $4w(w^2-1)=0$, so the branch points correspond to $w=0$ ($z=1$) and $w=\pm1$ ($z=0$).