$f(z)$ is Analytic then $f(\bar z)$ is analytic iff $f$ is constant.

Question

let $f_1=u(x,y), f_2=u(x,-y)$ Then is this true that

$(f_2)_x$ at point $(x,y) =(f_1)_x $ at point $(x,-y)$

And

$(f_2)_y $ at point $(x,y) = - (f_1)_y$ at point $(x,-y)$

If yes does this imply if $f(z)$ is Analytic then $f(\bar z)$ is analytic iff $f$ is constant.

Thanks in Advance.

Answer 1

Let be $z_0$ s.t. $f'(z_0)\ne 0$. Then $f$ is locally invertible with analytic inverse (why?). Then, near $z_0:$ $$z\longmapsto f(\bar z)\longmapsto f^{-1}(f(\bar z)) = \bar z$$ would be analytic. Contradiction.

Answer 2

Without CRD:

Let $g(z):= f(\overline{z})$ and show that

$\lim_{h \to 0 , h \in \mathbb R}\frac{g(z_0+h)-g(z_0)}{h}= f'(\overline{z_0})$

and

$\lim_{h \to 0 , h \in i\mathbb R}\frac{g(z_0+h)-g(z_0)}{h}= -f'(\overline{z_0})$.

Conclusion ?