I have the next problem:
Let f = u + iv be holomorphic in the region G and satisfy $u = h\circ v$ for some differentiable function $h : \mathbb{R} \longrightarrow \mathbb{R}$. Show that f is constant.
I did the following calculations:
If $f$ is holomorphic then: $u_x=v_y$ and $v_x=-u_y$. Then, $u_x=(h\circ v)_x=h_x(v)v_x$ and $u_y=(h\circ v)_y=h_y(v)v_y$. So, $v_x=-h_y(v)v_y=-h_yh_x(v)v_x$.
I think that $h_y=0$ because $h$ is real. But I'm not sure. I appreciate the help you can give me.
Best Answer
Partial answer: I will assume that $h$ is continuously differentiable.
$h$ is a function from $\mathbb R$ into itself so it has no partial derivatives.
We have $u_x=h'(v)v_x$ so $h'(v) v_x=v_y$. Also $u_y=h'(v)v_y$ so $h'(v)v_y=-v_x$. Consider the subset of $G$where $h'(v) \neq 0$. Suppose this is non-empty. This an open set. On this set we get $v_x=v_y=0$ so $v$ is a constant. Now C-R equation show that $u$ is also a constant so $f$ is constant. By the Identity Theorem this implies that $f$ is a constant throughout $G$. If our subset is empty the $h$ is a constant so $u$ is a constant an this makes $f$ a constant.