I am trying to solve the following problem.
$f(z)=u(x,y)+iv(x,y)$ is an analytic function in $D$ ($D$ is connected and open).
If $u, v$ fulfill the relation $G(u(x,y), v(x,y))= 0 $ in $D$ for some function ($G:\mathbb{R^2}\to\mathbb{R}$) with the property
$(\frac{\partial G}{\partial x})^2+(\frac{\partial G}{\partial y})^2 > 0$
Show that $f$ is constant.
My attempt:
We have to show that $f'(z)=0$
$(\frac{\partial G}{\partial x})^2+(\frac{\partial G}{\partial y})^2 = (\frac{\partial G}{\partial u})^2(\frac{\partial u}{\partial x})^2 +2\frac{\partial G}{\partial u}\frac{\partial u}{\partial x}\frac{\partial G}{\partial v}\frac{\partial v}{\partial x}+(\frac{\partial G}{\partial v})^2(\frac{\partial v}{\partial x})^2+(\frac{\partial G}{\partial u})^2(\frac{\partial u}{\partial y})^2 +2\frac{\partial G}{\partial u}\frac{\partial u}{\partial y}\frac{\partial G}{\partial v}\frac{\partial v}{\partial y}+(\frac{\partial G}{\partial v})^2(\frac{\partial v}{\partial y})^2$
Since $f$ is analytic, it satisfies the Cauchy Riemann equations, $ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$.
Then,
$(\frac{\partial G}{\partial x})^2+(\frac{\partial G}{\partial y})^2 = (\frac{\partial G}{\partial u})^2(\frac{\partial u}{\partial x})^2 +(\frac{\partial G}{\partial v})^2(\frac{\partial v}{\partial x})^2+(\frac{\partial G}{\partial u})^2(\frac{\partial u}{\partial y})^2 +(\frac{\partial G}{\partial v})^2(\frac{\partial v}{\partial y})^2 > 0$
But because $G(u(x,y), v(x,y))= 0 $ , doesn't this mean that $\frac{\partial G}{\partial u}=0 $ and $\frac{\partial G}{\partial v} = 0$? Then the inequality doesn't seem to make sense. What am I getting wrong? Is this the right approach?
Thanks in advance.
A better way is to try and visualize what's exactly going on behind the scenes.See, the condition in the problem implies that $G$ is differentiable with its total derivative $dG$ being a surjection from $\mathbb{C}$ to $\mathbb{R}$ at each point.Now since $G(f)=0$ we apply chain rule to get
$dG_{f(z)} \circ \hspace{1mm} df_{z} = 0$ for all $z$ in $D$.If for some $z$, $f'(z)$ is non-zero, then $df_{z}$ is a nontrivial scaled rotation, which is invertible. Hence $dG_{f(z)}(df_z(\mathbb{C}) = dG_{f(z)}(\mathbb{C}) = \mathbb{R}$, a contradiction.Hence for all $z$ in $D$ we have $f'(z)=0$.Since $D$ is open and connected, $f$ is constant.