Consider the complex function $$f(z) = \sqrt{z^2+1}.$$

Obviously, $f(z)$ has branch points at $z = \pm i$. One way of defining a branch cut would be to exclude the points on the imaginary axis with $|z| \geq 1$. Another way of defining a branch cut appears to be to exclude the (finite) region of the imaginary axis with $|z| \leq 1$.

If we define $f(z)$ as $$f(z) = e^{1/2\log(z^2+1)},$$
the first branch cut can be arrived at by taking the principal branch of $\log(z)$ with the branch cut $(-\infty,0]$. At first I thought the second branch cut could be arrived at by taking the branch cut $[0,\infty)$ for $\log(z)$. Indeed, this would exclude imaginary $z$ with $|z| \leq 1$, but it would of course also exclude all real $z$, thus constituting a different branch cut for $f(z)$.

I think it would be possible to arrive at the second branch cut for $f(z)$ differently, by defining $$f(z) = \sqrt{r\_1r\_2}e^{i(\theta\_1+\theta\_2)/2}, $$
where $r\_1 = |z-i|, r\_2 = |z+i|$, $\theta\_1 = \arg(z-i), \theta\_2 = \arg(z+i).$ However, I don't really like this approach, since I think $$f(z) = e^{1/2\log(z^2+1)}$$
is the proper way to define $f$.

Any comments on this? Is there a branch cut for $\log(z)$ which gives the correct branch cut for $f$?