[Math] If argument of $f$ is constant, then $f$ is constant

complex-analysis

$f$ is holomorphic on an open unit disk $D$. If $arg(f)$ is constant
in $D$, then $f$ is constant in $D$

I don't know where to start:

I know the Cauchy Riemann Equations (these must be satisfied for complex differentiability) :
$u_x=v_y$ and $v_x=-u_y$

I know the open mapping theorem – every open set is mapped to an open set

I know the maximum modulus theorem – if $f$ is non-constant, then it can't attain a maximum in $D$.

Best Answer

If the argument of $f$ is constant, writing $f$ as: $f = |f| \exp(\arg f)$, we see that $|f|$ must be holomorphic. Since $|f|$ has a constant imaginary part, it must be itself constant.