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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.

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up vote 2 down vote accepted

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 $\log(f(0))$.

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so to clarify as far as I can understand this the idea is that the value of the integral over the curve C is log(f(z)) - log(f(0))... how does this get around the issue of which branch of log is chosen? this seems equivalent to me to just defining a value of log, but I guess the argument works bc the integral of a holomorphic function is holomorphic, thus h is holomorphic? is the intuition because $e^{z}$ is a periodic function so we don't even have to take a branch, we can just define log along the branch cut and when we take $e$ to this power the resulting function is holomorphic? – Red Rover May 10 '12 at 19:35
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 $f(z)$. – Robert Israel May 10 '12 at 20:43

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