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In a book, there is a proposition states that : Let $U$ be an open set of $\mathbb{C}$ and $f$ be a holomorphic function on $U$ does not take value $0$. Then $\ln |f|$ is harmonic.

In the proof, it's written $\frac{\partial}{\partial z} \ln|f| = \frac{\overline{f}f'}{2\overline{f}f}$. But I don't know how this calculation be down. Can somone detail this？ Thanks.

My attempts is as follows : $\frac{\partial}{\partial z} \ln|f| = \frac{1}{4}(\frac{\partial \ln f\overline{f}}{\partial x} + i\frac{\partial \ln f\overline{f}}{\partial y}) = \frac{1}{4f\overline{f}} (f\frac{\partial\overline{f}}{\partial x} + \overline{f}\frac{\partial f}{\partial x} + if\frac{\partial\overline{f}}{\partial y} + i\overline{f}\frac{\partial f}{\partial y})$

ZENG
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1 Answers1

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Notice that $\ln \left|f\right|=\frac{1}{2}\ln f\overline{f}$ and $\frac{\partial \overline{f}}{\partial \overline{z}}=0$ since $f$ is holomorphic and thus locally we have the expansion $$f\left(z\right)=\sum_{n=0}^{\infty} \frac{f^{n}\left(z_0\right)}{n!}\left(z-z_0\right)^n$$ so $\overline{f}\left(z\right)$ would not have any term that entails $z$. Hence, we can see $$\frac{\partial \ln \left|f\right|}{\partial z}=\frac{1}{2}\frac{\partial \ln f\overline{f}}{\partial z}=\frac{1}{2f\overline{f}}\frac{\partial f\overline{f}}{\partial z}=\frac{\overline{f}f'}{2f\overline{f}}=\frac{f'}{2f}.$$ Now, if we do $\frac{\partial}{\partial \overline{z}}$, we have $$\frac{\partial^2 \ln \left|f\right|}{\partial \overline{z}\partial z}=\frac{\partial}{\partial \overline{z}} \frac{f'}{2f}=0$$ since $f,f'$ are both holomorphic. Hence, $\ln \left|f\right|$ is harmonic by the identity $$ \Delta=4\frac{\partial^2 }{\partial \overline{z}\partial z}.$$

Back to your attempt, it seems like it is pretty close to the desired form and you can get it if you realize $\frac{\partial \overline{f}}{\partial z}=0$ as mentioned above.

Eric
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