This question was asked in my complex analysis quiz.
Let f be holomorohic on U and assume that for each r $\in$ (0, 1) , f($re^{it} )$ has constant argument. Show that f must be constant.
The solutions of quiz are not given and I have no idea on which result I should use. I have done some questions in assignment related to proving a function to be constant but this is very different than them and I have no idea on which result to use.
Can you please help?
Best Answer
I'll assume that $f(z) \ne 0$ for $0 < |z| < 1$ so that the argument of $f(z)$ is well-defined. The idea is to show that $zf'(z)/f(z)$ is purely imaginary and therefore constant, and then derive a contradiction unless $f' \equiv 0$.
For fixed non-zero $z \in U$ and real $t > 0$ is it given that $f(ze^{it})$ and $f(z)$ have the same argument, so that $$ \frac{f(ze^{it}) - f(z)}{f(z)} $$ is a real number. Now we consider the identity $$ i \frac{f(ze^{it}) - f(z)}{ze^{it} - z} \cdot \frac{z}{f(z)} \cdot \frac{e^{it}-1}{it} = \frac{f(ze^{it}) - f(z)}{f(z)} \cdot \frac 1t \, . $$ For $t \to 0$ the left-hand side has the limit $$ i f'(z)\cdot \frac{z}{f(z)} \cdot 1 $$ and the right-hand side is a real number for all $t > 0$. It follows that $$ i\frac{zf'(z)}{f(z)} \in \Bbb R $$ for all non-zero $z \in U$. From the open mapping principle we can now conclude that the left-hand side is constant: $$ i\frac{zf'(z)}{f(z)} = k \in \Bbb R $$ for all non-zero $z$ in $U$. If $k=0$ then $f'$ is identically zero and we are done.
If $k \ne 0$ then necessarily $f(0) = 0$ with some multiplicity $p > 0$ and we get a contradiction from $$ p = \lim_{z \to 0} \frac{zf'(z)}{f(z)} = -ik \, . $$