Show that the holomorphic function $u + iv$ must be constant if $u = h \circ v$. Problem from Remmert. The problem as stated in R. Remmert's Theory of Complex Functions p. 62., is this:
Let $f = u + iv$ be holomorphic in the region $G \subseteq \mathbb{C}$ and satisfy $u = h \circ v$ for some differentiable function $h : \mathbb{R} \to \mathbb{R}$. Show that $f$ is constant.
I've attacked this computationally in each way I can think of. I've tried to show that $\frac{\partial f}{\partial z}$ is zero; played with the Laplace equation, knowing that $h \circ v$ and $v$ must be hamonically conjugate, etc. I even considered spinning an argument based on the maximum principle.
It seems obvious, but I have not succeeded in a proof yet. Perhaps I am missing a concept?
 A: We have 
 $$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}\cdot(h'\circ v)$$ 
and
 $$-\frac{\partial v}{\partial x}=\frac{\partial u}{\partial y}=\frac{\partial v}{\partial y}\cdot(h'\circ v).$$ 
Hence
$$ \frac{\partial v}{\partial y}\cdot (1+(h'\circ v)^2)=0$$
and the same with the other three partial derivatives. As the sum in parentheses is positive, we conclude tha tall partial  derivatives are zero, hence $f$ is constant. 
A: We have 
\begin{align*}
h'(v(x))\dfrac{\partial v}{\partial x}&=\dfrac{\partial v}{\partial y}\\
h'(v(x))\dfrac{\partial v}{\partial y}&=-\dfrac{\partial v}{\partial x},
\end{align*}
and hence
\begin{align*}
((h'(v(x)))^{2}+1)\dfrac{\partial v}{\partial y}=0.
\end{align*}
Likewise we have $\dfrac{\partial v}{\partial x}=0$.
A: I don't know what R. Remmert means by a region, but I think he means a connected open set; so I'll assume that's what $G$ is.  Then:
$z = x + iy; \tag 1$
since
$f(z) = u(x, y) + iv(x, y) \tag 1$
is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann equations,
$u_x = v_y, \tag 2$
$u_y = - v_x, \tag 3$
in $G$; note this implies
$\nabla u \cdot \nabla v = 0, \tag 4$
for
$\nabla u \cdot \nabla v = u_x v_x + u_y v_y = u_x(-u_y) + u_y u_x = -u_x u_y + u_y u_x = 0; \tag 5$
now with
$u = h \circ v, \tag 6$
we have, by the chain rule,
$u_x = h'(v)v_x, \; u_y = h'(v) v_y, \tag 7$
whence
$\nabla u = h'(v)\nabla v; \tag 8$
by (4), this yields
$\nabla u \cdot \nabla u = h'(v) \nabla v \cdot \nabla u = 0, \tag 9$
whence
$\nabla u = 0, \tag{10}$
or $u$ is constant on components of $G$.  Then by (2)-(3), we also have
$\nabla v = 0, \tag{11}$
and $v$ is also constant the components of $G$.  This shows $f = u + iv$ is constant on components, so if $G$ is connected, we are done.
