Let $f$ be holomorphic in $D(0;1)$. Show that $\overline{f(z)}$ is holomorhpic in $D(0;1)$ if and only if $f$ is constant.
It is clear to me that if $f$ is constant then $\overline{f(z)}$ is holomorphic since the Cauchy-Riemann equations will be satisfied. However, I'm not sure about how to show the other direction. How do I do that?
There are several ways to show that. The most elementary is probably to look at the Cauchy-Riemann equations again.
A little bit more advanced, but shorter would be:
If $\overline f$ is holomorphic, then so are $f+\overline f$ and $f-\overline f$. But these are $\mathbb R$-valued and $i\mathbb R$-valued functions respectively, hence not open mappings. Then they must both be constant. Now $f$ is the sum of constant functions.
If $f$ and $\overline f$ are holomorphic in a region $G$ (eg unit disc), then by Cauchy-Riemann, $$u_x = v_y, v_x = -u_y,u_x=-v_y,v_x=u_y \implies u_x=v_x=u_y=v_y=f_x=f_y=f'(z)=0$$
By Thm 2.17, $f$ is constant in $G$.