0

if $f : \mathbb{C}^* \to \mathbb{C}$ is holomorphic function such that $f(z) \neq 0$ $\forall z \in \mathbb{C}^*$. then is it true we can find $g$ holomorphic such that $$e^g = f$$

Usually if $f$ is non vanishing and its domain is simply connected then we can conclude the existence of such a $g$. However as $\mathbb{C}^*$ is not simply connected does that imply the log of this function is not well defined?

Bumblebee
  • 1
yasrey1710
  • 392
  • 2
    See for example https://math.stackexchange.com/q/2235204/42969 – Martin R Jan 15 '22 at 10:04
  • Iff $\int_{|z|=1}\frac{f'(z)}{f(z)}dz=0$, this is what you need for the primitive $g(z)=\log f(1)+\int_1^z \frac{f'(s)}{f(s)}ds$ to be well-defined (independent on the path $1\to z\subset \Bbb{C}^*$) – reuns Jan 15 '22 at 11:30

1 Answers1

1

It is not true. As an example take $f(z) = z$. If $e^{g(z)} = z$, then $g$ would be a holomorphic logarithm on $\mathbb C^*$. This does not exist.

Paul Frost
  • 87,968