TITLE: Show the level curves of $\log|f(z)|$ are orthogonal to those of $\operatorname{arg}(f(z))$.
QUESTION [5 upvotes]: 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 $\log|f(z)|$?

REPLY [2 votes]: 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 $\dfrac{\pi}{2(n+1)}$.