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Given a function $f(z) \in C$ that is analytic, prove that $g(z) = \overline{f(\bar z)}$ is analytic in the set $\{\bar z : z \in C \}$. This is for homework: tips would be appreciated.

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  • $\begingroup$ $\{\bar z : z \in \Bbb C \}=\Bbb C$, take $f(z)=z$, this is not true. $\endgroup$ – Hamou Oct 3 '14 at 2:32
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    $\begingroup$ The identity function is entire, but the complex conjugate function isn't. $\endgroup$ – Yuval Filmus Oct 3 '14 at 2:32
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    $\begingroup$ Maybe $g(z)=\overline{f(\bar z)}$? $\endgroup$ – Hamou Oct 3 '14 at 2:34
  • $\begingroup$ @Hamou Yes, that is correct. I misread the problem because the type was so small. $\endgroup$ – user2049004 Oct 3 '14 at 2:39
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$f(z)=\sum a_nz^n$ analytic ,hence $g(z)=\sum \bar a_n z^n$ is analytic.

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