Show that $f(z) = \overline{w(\bar{z})}$ is an analytic function, where $w(z)$ is analytic in specific domain

Suppose $w(z)$ is an analytic function in a domain $G$ which is symmetric with respect to the real axis. Show that $f(z) = \overline{w(\bar{z})}$ is then an analytic function of $z$ in $G$.

So this is interesting because we are looking at the conjugate of a function of a conjugate. Let $w(z) = u(x,y)+iv(x,y)$. Then $f(z) = \overline{u(x,y)-iv(x,y)}$. So then just see that the Cauchy-Riemann equations are satisfied? But this is a necessary condition. So we need a sufficient condition to show analyticity right?

• $w(z)=u(x)+iv(y)$ is incorrect, and $f(z)=\overline{u(x)-iv(y)}$ wouldn't follow if it were correct. C-R is also sufficient for $C^2$ functions; $w$ is $C^\infty$ so there is no problem. Dec 22 '10 at 21:26
• Isn't $w(z)$ just an arbitrary function? Dec 22 '10 at 21:31
• You need $w(z) = u(x,y) + i v(x,y)$. Dec 22 '10 at 21:37
• No. To quote the beginning of your question, "Suppose $w(z)$ is an analytic function". Dec 22 '10 at 21:37
• Your expression for $f$ is still incorrect. Note that $\overline{u-iv}=u+iv$, so you have the assertion that $f=w$. If $u(z)=u(x,y)$, then $u(\overline{z})=...$. Dec 22 '10 at 23:40