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Let G be a region and define $G^∗ = \{z : z̅ ∈ G\}$. If $f : G \to \mathbb{C}$ is analytic, prove that $f^* : G^∗ → C$, defined by $f^*(z) = \overline{f(\overline{z})}$ is also analytic.

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    $\begingroup$ You are missing a conjugate in the definition of $f*$ $\endgroup$ – mrf Sep 18 '15 at 19:05
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Another solution, different to the ones on the duplicates, is by using the Wirtinger Operators:

Let $f^*(z)=\overline{f(\overline{z})}$. Now since $f$ is analytic in $G$ iff $$\frac{\partial}{\partial \overline{z}} \ f=0$$ in $G$, and in this case one can see that the Cauchy Riemann equations for $f$ gives that also $$\frac{\partial}{\partial \overline{z}}f^*=0$$ in $G^*$, thus indeed $f^*$ is analytic in $G^*$.

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