The complex multifunction i^z and its "branches"

So I was looking at $i^z$, obviously this is a multi-valued function (a multifunction) due to the fact that $i^z=e^{z \log{i}}$ and $\log{z}$ is a multifunction. Now the usual multifunctions I know either have a finite or infinite number of branches at ALL points, except, of course, the singularities and the branch points.

For example: $\sqrt{1}=1 \text{ or } -1$ depending on if we take $1 \text{ to be } e^{i(0)} \text{ or } e^{i(2\pi)}$ respectively (or any other odd or even multiples of $2\pi$). And this applies to all the other complex numbers, except zero. And the same applies to the logarithm function, except here we have infinite solutions differing by an integer multiple of $2\pi i$.

However, when we come to look at $w=i^z$, we find that not all points in $z$ map to multiple points in $w$! Example: $i^1=i$ only, and $i^{\frac{1}{2}}=e^{i \frac{\pi}{4}} \text{ or } -e^{i \frac{\pi}{4}}$. As we can see here not all points in the z-plane will have the same number of corresponding points in the w-plane. So how many branches does this multifunction have? And what are the branch points?

Take a look at the Wikipedia article on "Exponentiation" under the subheading "complex exponents with complex bases". For $w^z$ the answer depend upon whether $z$ is integer, rational or irrational.

Hmmm so you're saying that this multifunction does not have any certain number of branches unless i restrict the domain to integer values? So would this be even considered a multifunction or a mapping for that matter?

What this tells you is that generalizing certain well-known functional forms from the real numbers to the complex numbers is not without its difficulties! In this case, if you fix a branch of the logarithm, then all is well; but if you look at the set of possible values of $f(z) = w^z$ as you range over the branches of the logarithm, then it's horribly behaved.

If you scroll down the page that I referenced to the section on "Failure of power and logarithm identities", you'll see some other problems that arise in this type of generalization.