# Prove that an analytic function with a constant imaginary part is constant itself

I want to prove that if $f$ is an analytic function for which $\Im f$ is constant, this implies that $f$ itself is constant.

So to start off, it's not given that the function is entire or anything, ruling out Liouville. Rather, I'm guessing the maximum modulus theorem will prove useful here (?).

If I were to show using that the real part of $f$ was constant, proving the statement would be easy (just take the absolute value and investigate $e^f$), but in this case I'm none the wiser from that. Should I use the same parametrization $e^f$ and apply the Cauchy-Riemann equations somewhere? Any suggestions welcome.

• If you want to pursue your own idea, look at $e^{if}$. – mrf Dec 25 '15 at 20:56

## 2 Answers

Hint: use the Cauchy-Riemann equations, $f(x,y)=g(x,y)+ih(x,y)$ if $h(x,y)$ is constant, $\partial_xg=\partial_yh=0$ and $\partial_yg=-\partial_xh=0$ thus $g$ is also constant. done.

• I guess this is the simplest proof.. – 1010011010 Dec 25 '15 at 17:58

The image of $f$ will be a horizontal line in $\mathbb{C}$. Now, consider the open mapping theorem.

• (Given that $f$ is defined on a "nice" domain, i.e. a simply connected region). – GiantTortoise1729 Dec 25 '15 at 17:50