$G\subset \Bbb{C}\setminus \{0\}$ be a domain (open and connected). Let $G^{\star}$ be the reflection of $G$ w.r.to the unit circle $S^1$. Given $f\in\mathcal{H}(G) $ ,then show that $f^{\star}\in\mathcal{H}(G^{\star})$ where $f^{\star}(z)=\overline{f(\frac{1}{\overline z}) }$.
My attempt:
A mobius map $T\in\text{Aut}(\mathbb{C}^{\infty})$ maps a straight line or circle into straight line or circle and being conformal it preserves angle between any two points. Hence $z^{\star}$ is the reflection of the point $z$ about $S$ (straight line or circle) then $T(z^{\star}) $ is the reflection of $T(z) $ about $T(S) $.
Consider the mobius map $T(z) =\frac{z-i}{z+i}$
Then $T$ maps Real line onto the unit circle. Since $\overline{z}$ is the reflection of $z$ w.r.to Real axis, $w^{\star}=T(\overline{z}) $ will be the reflection of $w=T(z) $ w.r.to the unit circle.
$w^{\star}=T(\overline{z}) =\frac{\overline{z}-i}{\overline{z}+i}=\frac{1}{\overline{\frac{z-i}{z+i}}}=\frac{1}{\overline{w}}$
$G^{\star}=\{\frac{1}{\overline{w}}:w\in G\}$
This proves that $f^{\star}$ is well defined on $G^{\star}$.
To show $f^{\star}\in\mathcal{H}(G^{\star}) $ , let $z_0\in G^{\star}$ .
Then $\frac{1}{\overline{z_0}}\in G$ and $f\in \mathcal{H}(G) $ implies $f$ has a convergent power series $B_{\frac{1}{\overline{z_0}}}(R) $ for some $R>0$ . Let $f(z) =\sum_{n\ge 0} a_n (z-\frac{1}{\overline{z_0}})^n$
Then $f^{\star}(z) =\sum_{n\ge 0}\overline{a_n} (\frac{1}{z}-\frac{1}{z_0})^n$
I want to express $f^{\star}$ in the form : $f^{\star}(z) =\sum_{n\ge 0}b_n(z-z_0)^n$
How to use $B_{\frac{1}{\overline{z_0}}}(R) $ to get the above form?