Nowhere zero holomorphic function on simply connected domain admits holomorphic logarithm

Definition

Suppose $U$ is a simply connected domain in $\mathbb {C}$: in other words $U$ is an open subset of $\mathbb {C}$ such that every path in $U$ is nullhomotopic. Suppose $f:U\to \mathbb {C}$ is a holomorphic function with the property that $f(z)\neq 0$ for $z\in U$. Then, there exists a function $g:U\to \mathbb {C}$ such that:

$e^{g(z)}=f(z)\ \forall \ z\in U$

Facts used

• Homotopy-invariance formulation of Cauchy's theorem: This essentially states that integating a holomorphic function along a nullhomotopic loop gives zero. In particular, this means that integrating a holomorphic function along any loop in a simply connected domain, gives zero.

Proof

Defining the candidate

Pick a point $z_{0}\in U$ and choose $g(z_{0})$ such that $e^{g(z_{0})}=f(z_{0})$. This can be done because $f(z_{0})\neq 0$.

Now, for any point $z\in U$, pick a path $\gamma$ from $z_{0}$ to $z$ and define:

$g(z):=g(z_{0})+\int _{\gamma }{\frac {f'(z)}{f(z)}}\,dz$

We first need to argue that $g(z)$ is well-defined and independent of the choice of $\gamma$. For this, it suffices to show that for any closed loop $\gamma$, we have:

$\int _{\gamma }{\frac {f'(z)}{f(z)}}\,dz=0$

The latter follows using the simple connectedness of $U$. Indeed, because of the simple connectedness of $U$, any closed curve inside $U$ separates it into two regions, and we can pick the inside region whose boundary is precisely $\gamma$. Then, applying the Goursat lemma and using the fact that since $f$ is holomorphic and nonzero, $f'/f$ is holomorphic, we get that the above integral is zero.

Thus $g$ is well-defined.

Proving that it works

This is a direct check that involves showing that $(e^{-g}f)'=0$.