Show that the holomorphic function $u + iv$ must be constant if $u = h \circ v$. Problem from Remmert.

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $\dfrac{\partial v}{\partial x}=0$.