let $g(z)=f(\overline z)$ ,is $g$ analytic in $\Omega$ ??,

Question

let $f$ is analytic in domain $\Omega$ in complex plane $\mathbb{C}$

let $g(z)=f(\overline z) \ \forall \ z \in \ \Omega $

is $g$ analytic in $\Omega$ ??

if not why??

i was trying it by C-R equations

let $$f(z)=u(x,y)+iv(x,y) $$ then$$g(z)=U(x,y)+iV(x,y)=u(x,-y)+iv(x,-y)$$ now does $$U_x=V_y,U_y=-V_x \ \ ??$$

Answer 1

You're working too generally to see what's going on. Try an example instead! What happens if $f(z) = z$? Is $g$ analytic then?

If you want to be general, use the chain rule to express $U_x, V_x, U_y$ and $V_y$ in terms of $u_x,u_y, v_x$ and $v_y$. Then use $u_x = v_y, u_y = -v_x$ to find out what relationship holds between $U_x, V_x, U_y$ and $V_y$.