Prove $f$ is constant in $\Omega$

Question

They ask me:

$\Omega$ is a domain and $f$ is holomorphic in $\Omega$, prove that if $\lvert f(z) \rvert$ is constant in $\Omega$ then $f$ is also constant in $\Omega$

My idea to prove this is use the maximum modulus principle like:

1. I say as $\lvert f(z) \rvert$ is constant then $\lvert f(z) \rvert = M$, $M$ arbitrary value
2. As $\lvert f(z) \rvert = M$ I noticed that $\lvert f(z) \rvert$ reach the maximum
3. So using the the maximum modulus principle I can affirm that $f$ is constant.

I don't know if it is correct or not.

Answer 1

Your proof is correct. My proof: suppose that $f$ is not constant. Then $f( \Omega)$ is open, but this is not possible, since $|f|$ is constant.

Answer 2

Your solution is correct. Note that this property is usually proved before proving the maximum modulus principle, using the Cauchy-Riemann equations:

$$|f(z)|=const\Rightarrow u^2(x,y)+v^2(x,y)=const$$

Taking the partial derivatives and using Cauchy-Riemann we get: $$2uu_x+2vv_x=0\Rightarrow2uu_x-2vu_y=0$$ $$2uu_y+2vv_y=0\Rightarrow2uu_y+2vu_x=0$$ We get a homogeneous system of 2 equations with variables $u_x,u_y$. For a non-trivial solution (which naturally gives a constant function since $u_x=u_y=0$ and similarly $v_x=v_y=0$), we required the determinant to be $0$ and so: $$4u^2+4v^2=0\Rightarrow u=v=0$$ Which once again gives a constant solution. Either way - $f$ is constant.