An holomorphic function with zero real or imaginary part is constant

Question

I'm trying to prove this proposition:

Let $f:\Omega \to \mathbb{C}$, $\Omega \subset \mathbb{C}$ a domain, such that $Re(f)=0$ or $Im(f)=0$. Prove that if $f$ is holomorphic on $\Omega$, then $f$ is constant on that domain.

So I see that it's true, because I know that holomorphic $\implies$ analytic $\implies$ $C^1(\Omega)$. So we can say that $\frac{\partial f}{\partial \bar{z}}=0$, and then use CR equations.

But I haven't been proved the first implication yet, so I can't use the fact that we know that $f$ is $C^1$.

How can we prove this then?

Thanks for your time.

Answer 1

Hint: Let $\Re{f} = u$ and $\Im{f} = v$, so that $f = u+iv$. If we assume that $f$ is holomorphic and that that $u = 0$, we get by the CR equations \begin{align} 0 &= \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\\ 0 &= -\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}. \end{align} Integrating the first equality w.r.t. $y$, we get $$ v = g(x) $$ for some function $g(x)$. What does the second equality tell us about $g$?

Answer 2

If $f=u+iv$ is holomorphic in $\Omega,$ then the CRE hold in $\Omega.$ If we also know $u\equiv 0$ in $\Omega,$ then both partials of $u$ vanish everywhere. Hence, by the CRE, both partials of $v$ vanish everwhere. Since we are in a connected open set, this implies $v$ is constant in $\Omega.$ Thus $f(z) = iC$ in $\Omega$ for some real constant $C.$