Show that if $f$ is analytic then $g(z)=\overline{f(\overline{z})}$ is analytic and find its derivative

Question

Let $f:\mathbb{C}\to\mathbb{C}$ be an analytic function. Show that $g(z)=\overline{f(\overline{z})}$ is analytic and find its derivative.

$f$ is analytic then C-R equations hold. Suppose $f=u+iv$.

Then $g(z)=u(x,-y)+i(-v(x,-y))$. We can observe that C-R equations hold for $g$ as well since if we set $\hat{u}(x,y):=u(x,-y)$ and $\hat{v}(x,y)=-v(x,-y)$ we get $$ \hat{u}_x=u_x=-v_y=-\hat{v}_y \\-\hat{u}_y=u_y=v_x=-v_x $$ But this is not enough for to show that $g$ is analytic.

Answer 1

If $f(z) = \sum\limits_{k=0}^{\infty} a_nz^{n}$ then $g(z)=\sum\limits_{k=0}^{\infty} \overline {a}_n z^{n}$. You can differentiate the series for $g$ term by term.

$g'$ is obtained from $f'$ the same way $g$ is obatined from $f$.