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Question:

Suppose $f(z)$ is analytic in a neighbourhood where $f(z) \neq 0$. Show the level curves of $\log|f(z)|$ are orthogonal to those of $\operatorname{arg}(f(z))$.

Here, I know I should use some of the angle preserving properties of conformal maps although I'm not sure how to begin. First, how do we characterize the level curves of $\log|f(z)|$?

mathjacks
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  • Look at the function $z\mapsto \log f(z)$ [any branch, locally]. What do you know about it? – Daniel Fischer Dec 08 '14 at 18:56
  • If we write $f(z) = re^{i\theta}$, then we know that the $\log f(z)$ depends on the size of $r$ and it's angle is defined by $\theta$. Perhaps I'm on the wrong track, I'm not sure what that helps achieve. – mathjacks Dec 08 '14 at 19:09

1 Answers1

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The level curves of $\log|f(z)|$ are the same as the level curves of $|f(z)|$. Now, a set of the form $\{z:|f(z)|=\epsilon\}$ (for some fixed $\epsilon>0$) is the pre-image of the circle $\{w:|w|=\epsilon\}$, while a set of the form $\{z:\arg(f(z))=\alpha\}$ (for some fixed $\alpha\in[0,2\pi)$) is the pre-image of the ray $\{w:\arg(w)=\alpha\}$.

Since circles and rays are perpendicular to each other in $w$-space, it follows that their pre-images under $f$ will be perpendicular in $z$-space anywhere that $f$ is conformal (ie angle preserving). Thus your original problem should also require that $f'\neq0$ as well.

At a zero of $f'$ of multiplicity $n$, the ramification of the angles is $n+1$, so the angle between the level curve of $|f|$ and of $\arg(f)$ is $\dfrac{\pi}{2(n+1)}$.

Trevor J Richards
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