Where is this function holomorphic?

Question

I've never really had to think about this problem before, and to be honest complex analysis isn't my strongest suit, so when I suddenly needed to know where $z^z$ is holomorphic, I didn't know where to begin to start proving my hunches. My guess is that it's holomorphic away from $0$.

More generally, given two entire functions $f$ and $g$, where is $f(z)^{g(z)}$ holomorphic? Again, my hunch is that it is away from the zeroes of $f$. This isn't a time sensitive question, so any help you can provide would be appreciated, from proofs to thoughts to hints!

$f(z)^{g(z)}=\exp [g(z) \log f(z)]$ if $f(z)\ne 0$ – luka5z Dec 17 '15 at 17:17 — luka5z, Dec 17 '15 at 17:17
That function is not single-valued. – Giuseppe Negro Dec 17 '15 at 17:45 — Giuseppe Negro, Dec 17 '15 at 17:45

score 3 · Answer 1 · answered Dec 17 '15 at 17:43

3

By definition, $z^z = \exp(z\log z)$ and to get a well-defined function, you need to choose a particular branch of $\log z$, and $z^z$ will be holomorphic wherever that branch of $\log$ is. With the most common choice (the so-called principal branch), $\log z$ is defined and holomorphic on the whole complex plane except the negative real axis ($0$ included).

Note that it is impossible to pick a branch of $\log$ that is holomorphic on $\mathbb{C} \setminus \{ 0 \}$.

answered Dec 17 '15 at 17:43

mrf

43,639

So is it also the case in a more general case? Domain of $f(z)^{g(z)}$ is the same as of $\log{f(z)}$? Or do we need to put some additional restrictions for this to be true? – mavzolej Feb 15 '20 at 06:29

Where is this function holomorphic?

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