Suppose that a smooth function $f=\left(f_{1}, f_{2}\right): \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ satisfies the following property:
- $$ \frac{\partial f_{1}}{\partial x}=\frac{\partial f_{2}}{\partial y}, \quad \frac{\partial f_{1}}{\partial y}=-\frac{\partial f_{2}}{\partial x} $$ Prove that, if the Jacobian matrix $D f(x, y) \neq 0,$ then $f$ is locally invertible and the inverse function also satisfies property (1).
My try: $f$ is locally invertible by Inverse function theorem. Consider $f (x, y)=\left(f_{1}(x, y), f_{2}(x, y)\right)=(z, \omega): \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ $\Rightarrow f^{-1}(z, \omega)=f^{-1} f(x, y)=(x, y)$ $\Rightarrow g(z, \omega)=(x, y).$ $\Rightarrow\left(g_{1}(z, \omega), g_{2}(z, \omega)\right)=(x, y).$ Observe that, if we consider, $$ \frac{\partial f_{1}}{\partial x} \neq 0 \neq \frac{\partial f_{1}}{\partial y} $$ Now we have,
$1=\frac{\partial g_{1}}{\partial x }=\frac{\partial g_{1}}{\partial z} \frac{\partial z}{\partial x}=\frac{\partial g_{1}}{\partial z} \frac{\partial f_{1}}{\partial x}$ $\Rightarrow \frac{\partial g_{1}}{\partial z}=\frac{1}{\left(\frac{\partial f_{1}}{\partial x}\right)}=\frac{1}{\left(\frac{\partial f_{2}}{\partial y}\right)}$
and again
$1=\frac{\partial g_{2}}{\partial y}=\frac{\partial g_{2}}{\partial \omega} \frac{\partial \omega}{\partial y}=\frac{\partial g_{2}}{\partial \omega} \frac{\partial f_{2}}{\partial y} \Rightarrow \frac{\partial g_{2}}{\partial \omega}=\frac{1}{\left(\frac{\partial f_{2}}{\partial y}\right)}.$
$\therefore$ we have,
$$\frac{\partial g_{1}}{\partial z}=\frac{\partial g_{2}}{\partial \omega}.$$
$1=\frac{\partial g_{1}}{\partial x}= \frac{\partial g_{1}}{\partial \omega}\frac{\partial \omega}{\partial x}=\frac{\partial g_{1}}{\partial \omega} \frac{\partial \omega}{\partial x}=\frac{\partial g_{1}}{\partial \omega} \frac{\partial f_{2}}{\partial x}\Rightarrow \frac{\partial g_{1}}{\partial \omega}=\frac{1}{\left(\frac{\partial f_{2}}{\partial x}\right)}=-\frac{1}{\left(\frac{\partial f_{1}}{\partial y}\right)}.$
$1=\frac{\partial g_{2}}{\partial y}=\frac{\partial g_{2}}{\partial z} \frac{\partial z}{\partial y}=\frac{\partial g_{2}}{\partial z} \frac{\partial f_{1}}{\partial y}.$
$\Rightarrow \frac{\partial g_{2}}{\partial y z}=\frac{1}{\left(\frac{\partial f_{i}}{\partial y}\right)} $ So, we have $\frac{\partial g_{1}}{\partial \omega}=-\frac{\partial g_{2}}{\partial z}$
What will happen if either of $\frac{\partial f_{1}}{\partial x}$ or $\frac{\partial f_{1}}{\partial y} $ is zero? I don't think that here we are allowed to use complex analysis because it was one of our advanced calculus question where it is assumed that we don't now complex analysis.