I am doing a problem concerning the branch points of the complex function $$f(z)=\log\left( \frac{1+\sqrt{1+z^2}}{2}\right)$$
The question begins by asking what the branch points of $f(z)$ are, and then tells you what they are; it says that $z=0,\pm i, \infty$ are the branch points. The latter three ($-i,i,\infty$) all make perfect sense, but the $z=0$ does not. Nor does Wolfram Alpha seem to think so either.
The problem continues by saying the following:
If we define $\sqrt{1+z^2}|_{z=0}=1$, show that the origin is not a branch point of this function. Draw a set of branch cuts to make the function single valued.
This makes it seem like it's saying that $z=0$ is a branch point of $\sqrt{1+z^2}$ by saying that $\sqrt{1+z^2}$ is multi-valued for positive real numbers. While $\sqrt{1+z^2}$ is certainly multi-valued, the branch points seem to simply be $i$ and $-i$, since these are the zeroes of $\sqrt{1+z^2}$, and I don't see how $\sqrt{1+z^2}$ is multi-valued at $z=0$ (though theoretically $-1$ is another root of 1).
I would be very grateful if someone could give me a hint as to why 0 is a branch point of this function and why the definition of $\sqrt{1+z^2}$ at $z=0$ is necessary.
We must dispense the assumption that the square root of a positive integer is uniquely defined; indeed, $(-1)^2=1^2=1$, so that both $-1$ and 1 are valid roots of $1$.
Theoretically, this is an issue for every positive integer, but the reason that $\sqrt{1}=-1$ is important is that it offers the opportunity for $\frac{1+\sqrt{1+z^2}}{2}$ to be zero, at which point $\log \frac{1+\sqrt{1+z^2}}{2}$ will be undefined, and will become a branch point of $\log$. By defining $\sqrt{1+z^2}$ at $z=0$ to be $1$, we eliminate the possibility of there being a zero of $\frac{1+\sqrt{1+z^2}}{2}$, and thus $\infty,\pm i$ are the only remaining branch points.