Where is $f(z) = z^z$ holomorphic?

Question

I've been messing around with this complex function visualizer: davidbau.com/conformal/

The picture it shows for $f(z) = z^z$ is pretty crazy, so I was wondering: is this function holomorphic?

In the real case we'd find the derivative of $f(x) = x^x$ by taking the log of both sides and applying the chain rule. Can we still do that here?

I think that it is not holomorphic. At $z=0$ is not defined so it cannot be holomorphic on $z\in \mathbb{C}$. — MrYouMath, Oct 25 '17 at 14:48
Do you mean that it's not holomorphic anywhere in $\mathbb{C}$, or just that it's not entire? — Sambo, Oct 25 '17 at 14:49
Pretty sure they meant entire. A function can be undefined at $z=0$ and still be analytic in some open subset of $\mathbb{C}$. — Jürgen Sukumaran, Oct 25 '17 at 14:52

score 5 · Accepted Answer · answered Oct 25 '17 at 15:09

5

$z^z = \exp(z\log z)$ is multi-valued. Choosing any specific branch of $\log$ gives a branch of $z^z$ which is holomorphic where $\log$ is.

For example, taking $\log$ as the principal branch, we get a $z^z$ which is holomorphic everywhere except at the negative real axis.

answered Oct 25 '17 at 15:09

mrf

Indeed: https://math.stackexchange.com/a/1580070/42969 :) – Martin R Oct 25 '17 at 17:48
@MartinR Good find! Looks like I'm getting senile and can't remember what I did two years ago! – mrf Oct 25 '17 at 20:22

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