Show that if $f$ is complex differentiable and in some region $U$ it is $f'=0$, then $f$ is constant I have to show show that if $f$ is complex differentiable and in some region $U$ it is $f'=0$, then $f$ is constant. How can one prove it?

Let $a\in U$ and $V=\{z\in U / f(z) =f(a)\}$. Show that $V$ is open and closed in $U$.
Let $b $ in $ V$ hence $ b$ in $ U$. There is $ r>0$ st the ball $B(b,r)$ is in $U$.Fix $z$ in this ball. The map $g(t)=f(tb+(1-t)z)$ is dif and $g'=0$ then $g(0)=g(1)$. This show the opnes.
$V$ is closed in $U$ because $f$ is continuous.
