Given that $f(z)=u(x,y)+iv(x,y)$ and Cauchy-Riemann equations hold for $f$ at a point $z_0=x_0+iy_0$ what can be said about $g(z)=\overline{f(\overline{z})}=u(x,-y)-iv(x,-y)$ at the same point?
I tried to solve that using the chain rule defining $w=-y$ but it leads to a wrong answer.
$$\frac{\partial{g}}{\partial{\overline{z}}}=\frac{1}{2}\bigg(\frac{\partial{g}}{\partial{x}}+i\frac{\partial{g}}{\partial{y}}\bigg)=0$$
$$\frac{\partial{g}}{\partial{x}}=\frac{\partial{g}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{g}}{\partial{v}}\frac{\partial{v}}{\partial{x}}=\frac{\partial{u}}{\partial{x}}-i\frac{\partial{v}}{\partial{x}}$$
$$\frac{\partial{g}}{\partial{y}}=\frac{\partial{g}}{\partial{u}}\frac{\partial{u}}{\partial{w}}\frac{\partial{w}}{\partial{y}}+\frac{\partial{g}}{\partial{v}}\frac{\partial{v}}{\partial{w}}\frac{\partial{w}}{\partial{y}}=-\frac{\partial{u}}{\partial{w}}+i\frac{\partial{v}}{\partial{w}}$$
$$2\frac{\partial{g}}{\partial{\overline{z}}}=\bigg(\frac{\partial{u}}{\partial{x}}-\frac{\partial{v}}{\partial{w}}\bigg)-i\bigg(\frac{\partial{v}}{\partial{x}}+\frac{\partial{u}}{\partial{w}}\bigg)=0$$
Since Cauchy-Riemann hold for $f$, $\frac{\partial{u}}{\partial{x}}=\frac{\partial{v}}{\partial{y}}$ and $\frac{\partial{v}}{\partial{x}}=-\frac{\partial{u}}{\partial{y}}$:
$$\frac{\partial{v}}{\partial{y}}=\frac{\partial{v}}{\partial{w}}$$
$$\frac{\partial{u}}{\partial{y}}=\frac{\partial{u}}{\partial{w}}$$
That implies Cauchy Riemann equations don't hold for $g$ at $z_0$ since $w=-y$ which is wrong. I would like to know why the above approach is failing.
I read these related questions but they don't light me up.
let f be a nonconstant analytic function in the domain D, show that the function $g(z)= \overline {f(z)}$ is not analytic in D
Show that $\overline{f(\overline{z})}$ is holomorphic on the domain $D^*:=\{\overline z: z\in D\}$ using Cauchy Riemann equation.
$f(z)$ and $\overline{f(\overline{z})}$ simultaneously holomorphic
