# Prove that there is holomorphic function $f:G \rightarrow \mathbb{C}$ with $e^{f(z)}=z$

Let $G \subset \mathbb{C}\setminus\{0\}$ be a simply connected domain. Prove that there is a holomorphic function $f:G \rightarrow \mathbb{C}$ with $e^{f(z)}=z$ for all $z\in G$ and furthermore $\{f+2 \pi i: k\in \mathbb{Z}\}$ is the set of all holomorphic functions on $G$ with this property.

My approach to this exercise was to use a corollary which we discussed in a lecture:

Corollary: Let $G$ be simply connected and $K \subset G$ be an open disc. Is a holomorphic function $f_0:K \rightarrow \mathbb{C}$ analytically expendable along every path in $G$, then $f_0$ is the restriction of exactly one holomorphic function $f$ on $G$.

I mean the result to this exercise will be the the complex logarithm and all its infinite branches. But is there a solution without this knowledge. I guess the exercise is meant to show the existence without an explicit function.

Some help would be nice!

• In the title "homomorphic" should be "holomorphic"? – hardmath May 7 '18 at 23:38
• I don’t think it is true unless $G$ is simply connected. – Thomas Andrews May 7 '18 at 23:45
• I'm with @ThomasAndrews on this one. If anyone of youse guys can find such an $f(z)$, I will eat my holomorphic hat! – Robert Lewis May 8 '18 at 0:47
• Sorry for my mistakes describing the problem. I guess I got confused translating the statment from german to english. I edited the Question now. – thehardyreader May 8 '18 at 7:38
• @ThomasAndrews Can you give me a hint for the edited version of my question? – thehardyreader May 8 '18 at 11:24

The statement in your question is not true. We need the following stronger condition: Every connected component in $G$ is simply connected. If we have this, without loss of generality let $G$ be connected, then pick an arbitrary point $z_0\in G$, a value of $f(z_0)$ satisfying $e^{f(z_0)}=z_0$, and define $f(z)=f(z_0)+\int_{z_0}^z\frac{1}{w}dw$, which is well-defined since every loop is homotopic to a point.
To see the condition is necessary, let $G=\mathbb{C}\setminus\lbrace0\rbrace$, and suppose there exists such an $f$. Then integrate $f'(z)=\frac{{e^{f(z)}}'}{e^{f(z)}}=\frac{z'}{z}=\frac{1}{z}$ about the unit circle, getting $0=2i\pi$, a contradiction.