# When $|f|$ and $\arg(f)$ are analytic?

Let $f:ℂ→ℂ$ be an analytic function. Define $|f|$ and $\arg(f)$ be the modulus and the argument of $f$. Generally, $|f|$ and $\arg(f)$ are not analytic.

My question is about the cases where this happen: $|f|$ and $\arg(f)$ are analytic.

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Only if $f$ is constant en.wikipedia.org/wiki/Picard_theorem – jspecter Dec 30 '12 at 15:58
But there exists some cases where $|f|$ or $arg(f)$ are analytic. – ZE1 Dec 30 '12 at 16:09
@RH1: If even one of them is analytic it must be because $f$ is constant. – Henning Makholm Dec 30 '12 at 16:13
Ok and thank you. – ZE1 Dec 30 '12 at 16:18
I changed $arg(f)$, coded as arg(f), to $\arg(f)$, coded as \arg(f). – Michael Hardy Dec 30 '12 at 16:43

The values of a nonconstant analytic function defined on some domain $\Omega\subset{\mathbb C}$ always fill another two-dimensional domain $\Omega'\subset{\mathbb C}$. Given an analytic $f:\ \Omega\to{\mathbb C}$ the two functions $|f|$ and $\arg(f)$ are real-valued, resp. $S^1$-valued; therefore they cannot be analytic.
There is, however, a way to introduce these two functions into the analytic realm: Assume that $\Omega$ is simply connected and that $f(z)\ne0$ for all $z\in\Omega$. Then there is an analytic function $z\mapsto g(z)$ such that $$f(z)=e^{g(z)}\quad (z\in\Omega)\ .$$ This function $g$ can be considered as the logarithm of $f$. Going through the details you will see that in fact $$g(z)=\log\bigl|f(z)\bigr| + i \arg\bigl(f(z)\bigr)\quad (z\in\Omega)$$ (note that $g$ is only determined up to an additive multiple of $2\pi i$). So you can say that $\arg\bigl(f(z)\bigr)$ is the imaginary part of $\log\bigl(f(z)\bigr)$.