An analytic function $f$ with constant argument is a constant function. In Ahlfors textbook (p.72 of the 3rd edition) is the proof of the statement:
An analytic function $f$ in a region $\Omega$ whose argument $arg(f)$ is constant must be a constant function.
First he proves that if $Re(f)$ is a constant function then $f$ is constant. I've understood his proof for that. And now I quote his proof for the original statement:
"... if $arg(f)$ is constant, we can set $u=kv$ $\space$with constant $k$ (unless $v$ is identically zero). But $u-kv$ is the real part of $(1+ik)f$, and we conclude again that $f$ must be reduce to a constant."
I can't see why this shows that $f$ is constant, I would appreciate if someone could explain me this proof.


But $u - kv$ is the real part of $(1+ik)f$

and $u-kv$ is constant [unless $v \equiv 0$, when we observe that $f$ is real-valued, hence constant]. So by the fact proved before, $(1+ik)f$ is constant. But since $1+ik$ is a nonzero constant, that is equivalent to the constancy of $f$.
