We want to prove the following:

If $f$ is a holomorphic function on the unit disc $\mathbb{D}$ s.t. $f(z) \neq 0$ for $z \in \mathbb{D}$, then there is a holomorphic function $g$ on $\mathbb{D}$ such that $f(z)=e^{g(z)}$ for all $z \in \mathbb{D}$.

For such a $g$, we have that $\frac{f'(z)}{f(z)}=g'(z)$ on $\mathbb{D}$ since $f$ has non-zero derivative.

Perhaps then we define $g(z)=\int_\gamma \frac{f'(z)}{f(z)}dz$ where $\gamma$ is a path in $\mathbb{D}$ from $0$ to $z$.

Does this work?

  • $\begingroup$ Don't you need $f(z)$ not to be $0$? (as opposed to the derivative). $\endgroup$ – Jakub Konieczny Jun 11 '13 at 8:51
  • $\begingroup$ Sorry, I misread the text. Yes, I mean $f$ non-zero. I'll edit to that effect. $\endgroup$ – Mathmo Jun 11 '13 at 8:52
  • $\begingroup$ You probably mean that the function itself does not vanish, not its derivative. Otherwise, $f(z)=z$ would serve as a counterexample. $\endgroup$ – leshik Jun 11 '13 at 8:53

As far as I am able to tell, your idea is correct. You can define $g$ by: $$ g(z) = \int_\gamma \frac{f'(z)}{f(z)} dz$$ because $\mathbb{D}$ is simply connected and the integral does not depend on the path. Or even more simply, just expand $\frac{f'(z)}{f(z)} = \sum_{n} a_nz^n$ and put $g(z) = \sum_{n} \frac{a_n}{n+1} z^{n+1}$. In any case, you find a holomorphic $g$ such that $g'(z) = \frac{f'(z)}{f(z)}$.

Put $h(z) = e^{g(z)}$; this is again holomorphic. Then $\frac{h'(z)}{h(z)} = \frac{f'(z)}{f(z)}$ by a simple computation. Let $r(z) := \frac{f(z)}{h(z)}$; one can compute that: $$ r'(z) = \frac{f'(z)h(z) - h'(z)f(z)}{h(z)^2} = 0 $$ Hence, $r(z)$ is a constant function. As a consequence, $f(z) = c h(z)$ for some constant $c$. Writing $c = e^{\alpha}$ (which is always possible) you get: $$ f(z) = e^{\alpha + g(z)} $$ so $\tilde{g}(z) := \alpha + g(z)$ is the sought function.


Your idea is correct.

The function $g(z)$ you are trying to find can be think of as $\log f(z)$.

Note that $f \neq 0$, so $\frac{f'}{f}$ is a holomorphic function on the unit disk. Also the unit disk is simply connected, so any integration of $\frac{f'}{f}$ on a closed curve vanishes. So $g(z) = \int_0^z\frac{f'(\zeta)}{f(\zeta)}d\zeta$ is well-defined, which is a desired function.

You should notice that $\log f$ can be well-defined in more general cases. Suppose $f$ is a holomorphic function and $\Omega \subset \mathbb{C}$ is a simply connected region such that $f(z) \neq 0, \forall z \in \Omega$, then we can define $\log f(z) = \int_{z_0}^z\frac{f'(\zeta)}{f(\zeta)}d\zeta$, which is a well-defined holomorphic function on $\Omega$.


Yes, it does work.

In order to prove that in fact you have $f(z) = e^{g(z)}$, derivate $\frac{e^{g(z)}}{f(z)}$.

  • $\begingroup$ Well, $e^{g(0)} = 1$ while $f(0)$ can be anything nonzero, so you need to adjust things a little bit :-) $\endgroup$ – Martin Jun 11 '13 at 9:16
  • $\begingroup$ @Martin: Sure, I only wanted to give a hint on how to proceed, without spoiling the exercise more than necessary :) $\endgroup$ – Daniel Robert-Nicoud Jun 11 '13 at 16:02

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