Representation of Holomorphic Functions By Exponential

Question

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.

Answer 1

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 $\log(f(0))$.

Answer 2

Lef $f$ be holomorphic and non vanishing on the unitary disk. Prove $f(z)=e^{h(z)}$ for some holomorphic $h(z)$.

We play formally just for fun and then we turn this into a rigorous argument:

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Playing around: Suppose $f(z)=e^{h(z)}$, then $f'(z)=h'(z)e^{h(z)}=h'(z)f(z)$. Because $f(z)$ is nonzero we may divide to obtain:

$$\frac{f'(z)}{f(z)}=h'(z)$$

We notice that $f'(z)/f(z)$ is holomorphic so it must have an antiderivative in a simply connected domain as a corollary of Cauchy's Theorem, which we call $H(z)$.

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Rigorous Proof: Let $H(z)$ be as chosen in our initial discussion (an antiderivative of $f'/f$). In this case, we have that:

$$\frac{d}{dz}e^{-H(z)}f(z)=-H'(z)e^{-H(z)}f(z)+e^{-H(z)}f'(z)=-f'(z)e^{-H(z)}+e^{-H(z)}f'(z)=0$$

Hence, $e^{-H(z)}f(z)=c$ and we may write this as $c=e^{z_o}$ where $z_o=\ln|c|+i \arg c$. Thus, $f(z)=e^{z_o+H(z)}$ and $z_o+H(z)$ is analytic by our construction.

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This finishes the answer to the original post, but we wish to also ponder about a harder bonus question aswell, which is closely related to your problem:

Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be holomorphic on an annulus $P=\{z: r<|z|<R\}$ with finite zeros and poles in the disk $\{z: |z|\leq r\}$ and non vanishing on $P$, then $f(z)=z^n e^{h(z)}$ for some $n\in \mathbb{Z}$ and $h$ holomorphic on $P$.

We try to apply the same reasoning as above. If we differentiate $z^n e^{h(z)}=f(z)$ and isolate $h'(z)$ we have that $h'(z)=f'(z)/f(z)-n/z$ (of course this is only formal, but it guives us some guidance).

Let us make this rigorous. We have that by the argument principle that if $\gamma$ is a closed simple curve in $P$:

$$\int_\gamma \frac{f'(z)}{f(z)}dz =2\pi i(z-p)$$

Let $n=z-p$. In this case:

$$\int_\gamma \frac{f'(z)}{f(z)}-\frac{n}{z}dz=0$$

Because any closed curve in our domain is zero (even those that are not homotopic to zero), then we may define an anti-derivative for $f'(z)/f(z)-n/z$ given by integration and we call this anti-derivative $H(z)$. If $n=0$ we are in the previous case. Suppose not. We notice that:

$$\frac{d}{dz}e^{-H(z)}z^{-n}f(z)=-\left(\frac{f'(z)}{f(z)}-\frac{n}{z}\right)z^{-n}e^{-H(z)}f(z)$$ $$-ne^{-H(z)}z^{-n-1}f(z)+e^{-H(z)}z^{-n}f'(z)=0$$

Hence, we have that:

$$e^{-H(z)}z^{-n}f(z)=c$$ $$\therefore f(z)=e^{z_o+H(z)}z^n$$

Here is a final observation. We may drop the assumption there is a finite number of zeros and poles in $\{z: |z|<r\}$ and it still holds that $f(z)=z^n e^{h(z)}$ as may be seen here.