Show that if $f$ is analytic in a domain $D$ and $|f(z)|$ is constant in $D$, then the function $f(z)$ is constant in $D$

Question

Show that if $f$ is analytic in a domain $D$ and $|f(z)|$ is constant in $D$, then the function $f(z)$ is constant in $D$.

Here is what I try so far.

So we are given that $|f(z)|$ is constant in $D$, so that means $|f(z)|^2 = u^2 + v^2$ is also constant. If either $u = 0$, or $v = 0$, then $f(z)$ is constant.

$$\frac{\partial|f(z)|^2}{\partial x} = 2u \frac{\partial u}{\partial x} + 2v \frac{\partial v}{\partial x} = 0$$

$$\frac{\partial|f(z)|^2}{\partial y} = 2u \frac{\partial u}{\partial y} + 2v \frac{\partial v}{\partial y} = -2u \frac{\partial v}{\partial y} + 2v \frac{\partial u}{\partial x} = 0 $$

So what do I do to finish this proof? Would I have to use the Cauchy Riemann equation?

Answer 1

Assume that $c\ne 0$ is such that \begin{align*} |f(z)|&=c\\ |f(z)|^{2}&=c^{2}\\ f(z)\overline{f(z)}&=c^{2}\\ \dfrac{\partial}{\partial z}f(z)\overline{f(z)}&=0\\ \dfrac{\partial f}{\partial z}\cdot\overline{f(z)}+f(z)\cdot\overline{\dfrac{\partial f}{\partial\overline{z}}}&=0\\ \dfrac{\partial f}{\partial z}\cdot\overline{f(z)}&=0\\ \dfrac{\partial f}{\partial z}&=0, \end{align*} so $f$ is a constant.

Answer 2

Using $u_x$, $u_y$, $v_x$ and $v_y$ for the partial derivatives, you have \begin{cases} uu_x+vv_x=0 \\[4px] uu_y+vv_y=0 \end{cases} By C-R, $v_y=u_x$ and $v_x=-u_y$, so you have \begin{cases} uu_x-vu_y=0 \\[4px] vu_x+uu_y=0 \end{cases} Multiply the first equation by $u$, the second by $v$ and sum: $$ (u^2+v^2)u_x=0 $$ If $f$ is not the constant zero, the set where $u^2+v^2=0$ has no limit point. Thus its complement, where $u_x=0$, is dense in $D$. Hence $u_x=0$. Similarly, $u_y=0$, $v_x=0$ and $v_y=0$.