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Let $f:U\rightarrow\mathbb{C}$ be holomorphic on some open domain $U\subset\hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$ and $f(z)\not=0$ for $z\in U$.

Is it true that $z\mapsto \log(|f(z)|)$ is harmonic on $U$ ?

I guess the answer is yes and if that is true, how can I see that without a long and nasty calculation?

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2 Answers 2

up vote 4 down vote accepted

Yes, this is true. If $f(z)$ is holomorphic and nonzero on $U$, then $1/f(z)$ and $f'(z)$ are also holomorphic, so $f'(z)/f(z)$ is holomorphic on $U$. Therefore its integral $\log(|f(z)|) + i\arg(f(z))$ is holomorphic on any simply connected subdomain of $U$, so in particular $\log(|f(z)|)$ is harmonic everywhere on $U$.

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Not so fast. The derivative of $\log f(z)$ is $f'(z)/f(z)$, not $1/f(z)$. Where's the $f'(z)$? –  Robert Israel Oct 11 '12 at 17:15
    
Whoops, you're absolutely right. I'll fix that. –  Owen Biesel Oct 11 '12 at 17:20

Locally (but not necessarily globally), $\log f(z)$ is an analytic function because $\log$ is an analytic function. The real part of $\log f(z)$ is $\log |f(z)|$, i.e. the polar representation of the complex number $w$ is $w = r e^{i\theta}$ where $r = |w|$, and $\log w = \log r + i \theta$ so $\text{Re}(\log w) = \log r = \log |w|$. The real part of an analytic function is harmonic.

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