1
$\begingroup$

Let $f$ be holomorphic and nonzero on $D_{1}(0)$ the open unit disc. Can we write (for the given domain) $f(z) = e^{h(z)}$ where $h$ is holomorphic? This seems clear using a naive log argument but I'm having trouble with the issue of taking a branch of log.

1 Answers 1

3

Yes, this is true on any simply connected domain. In this case you can take $h(z) = h(0) + \int_C \dfrac{f'(\zeta)}{f(\zeta)}\ d\zeta = h(0) + \int_0^1 \dfrac{f'(tz)}{f(tz)} z\ dt $ where $C$ is the straight line from $0$ to $z$ and $h(0)$ is any branch of $\log(f(0))$.

  • 0
    The point is that $h(z)$ is an antiderivative of $f'(z)/f(z)$; on a simply-connected domain, any holomorphic function has a holomorphic antiderivative. From this you get $(f e^{-h})' = (f' - f h') e^{-h} = 0$, so $f/e^h$ is constant. There is no "branch cut" here: indeed, $h$ may well take different values at two points where $f$ has the same value, so it's not that we're fixing a particular branch of the logarithm and taking that function of $f(z)$.2012-05-10