If the absolute value of an analytic function is constant, then the function itself is constant. In the following answer: https://math.stackexchange.com/a/423148 it is proven that If the absolute value of an analytic function is constant, then the function itself is constant.
How does the result follow directly from the last step? (I know nothing about solving PDE's).
 A: The matrix $$
\left(\begin{array}{cc}
u & v \\ -v & u
\end{array}\right)
$$
is not singular if $u^2+v^2\neq 0$. So the equations 
$$
\left(\begin{array}{cc}
u & v \\ -v & u
\end{array}\right)\left(\begin{array}{c}v_y \\ v_x\end{array}\right)=\left(\begin{array}{c}0 \\ 0\end{array}\right)
$$
implies $v_x=v_y=0$. Then $v$ is constant.
A: One can aid the proof with a previous result. 

If $f$ is differentiable on an open disk $D(z_0)$ and if either $u$, $v$ or $|f|$ are constant on $D$, then so is $f$. If $f'=0$ on the disk, $f$ is also constant.

P Write $z_0=(a,b)$. Since $u$ is constant, Cauchy-Riemann gives $D_2v=D_1v=0$. By the usual one dimensional mean value theorem, $$v(x,y)-v(x,b)=D_2v(x,c')(y-b)=0$$ and $$v(x,b)-v(a,b)=D_1v(c'',b))(x-a)=0$$ so $v$ is constant. Analogously, we prove $u$ is constant provided $v$ is. 
If $u^2+v^2=k$ we obtain that $$\begin{cases}uD_1u+vD_1v=0\\uD_2u+vD_2v=0\end{cases}$$
Cauchy Riemann gives let's us write  $$\begin{cases}uD_1u+vD_1v=0\\uD_1v-vD_1u=0\end{cases}$$
which is $$\begin{pmatrix}u&v\\-v&u\end{pmatrix}\begin{pmatrix}D_1u\\D_1v\end{pmatrix}=0$$
If $|f|=0$, we're done. If not, we can invert the matrix to get $D_1u=D_1v=0$, so by Cauchy Riemann $D_2v=D_2u=0$. Can you continue like the first part of the proof?
