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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?

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$w(z)=u(x)+iv(y)$ 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. –  Jonas Meyer Dec 22 '10 at 21:26
    
Isn't $w(z)$ just an arbitrary function? –  PEV Dec 22 '10 at 21:31
    
You need $w(z) = u(x,y) + i v(x,y)$. –  Aryabhata Dec 22 '10 at 21:37
    
No. To quote the beginning of your question, "Suppose $w(z)$ is an analytic function". –  Jonas Meyer 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 $u(\overline{z})=...$. –  Jonas Meyer Dec 22 '10 at 23:40

1 Answer 1

up vote 4 down vote accepted

See Reflection Principle.

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