If $f$ is analytic, prove that $\overline{f(\overline{z})}$ is also analytic Let $f$ be an analytic function in an open set $U \subseteq \mathbb{C}$. Let $V=\{z\in\mathbb C:\overline z\in U\}$. Define $g$ on $V$ by $g(z)=\overline{f(\overline{z})}$. Show that $g$ is analytic on $V$. Note: It is not sufficient to show that $g$ is holomorphic because we haven't yet proved that holomorphic implies analytic.

Here is my attempted proof:
First, we will simply state what it means that $f$ is analytic on $U$. Applying the definition, we know that since $f$ is analytic on $U$, we know that for every $z_0 \in U$, there exists $r>0$ and a sequence of complex numbers $\left(a_n\right)_{n=0}^\infty$ such that $f(z)=\sum_{n=0}^\infty a_n\left(z-z_0\right)^n$ on the disc $D(z_0,r)$.
Next, note by definition of $V$ that for all $v \in V$, we have $\overline{v} \in U$. Therefore since $f$ is analytic on $U$, then for all $v \in V$ there exists $r>0$ and a sequence of complex numbers $\left(a_n\right)_{n=0}^\infty$ such that $f(u)=\sum_{n=0}^\infty a_n\left(u-\overline{v}\right)^n$ for all $u \in D(\overline{v},r)$.
Edit (Following Daniel Fischer's suggestion) Now if we realize that if $u \in D(\overline{v},r)$, then we see $u=\overline{w}$ for some $w \in D(v,r) \subseteq V$.
Therefore, putting everything together, we have that for all $v \in V$, there exists $r>0$ and a sequence of complex numbers $\left(a_n\right)_{n=0}^\infty$ such that $f(w)=\sum_{n=0}^\infty \overline{a_n}\left(w-v\right)^n$ for all $w \in D(v,r)\subseteq V$ (is this last part correct?).
I feel like I am getting close; I just need to obtain $g$ from this somehow. But I am losing direction here... Am I on the right track? Thanks!
 A: Splitting into real and imaginary parts probably isn't all that helpful - use the fact that $\overline{p(z)}=p(\bar{z})$ for every polynomial $p$ and the continuity of $z\mapsto\bar{z}$. Let $z_0\in U$. There exists $r>0$ and a sequence $(a_n)$ in $\mathbb{C}$ such that
$$f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k$$
for all $z\in\overline{D(z_0,r)}$. Define
$$f_n(z):=\sum_{k=0}^na_k(z-z_0)^k$$
for each $n\in\mathbb{N}$. Then $f_n\to f$ uniformly on $\overline{D(z_0,r)}$. We also define
$$g_n(z):=\overline{f_n(\bar{z})}=\sum_{k=0}^n\bar{a}_k(z-\bar{z}_0)^k$$
We have that $\liminf_{n\to\infty}|\bar{a}_n|^{-1/n}=\liminf_{n\to\infty}|a_n|^{-1/n}>r>0$ so the series
$$\sum_{k=0}^\infty\bar{a}_k(z-\bar{z}_0)^k$$
has radius of converge $\rho>r>0$. In particular, this implies that the above series defines an analytic function $g(z)$ on $D(\bar{z}_0,\rho)$ and $g_n\to g$ uniformly on $\overline{D(\bar{z}_0,r)}$. It follows from the continuity of $z\mapsto\bar{z}$ that $g(z)=\overline{f(\bar{z})}$ on $\overline{D(\bar{z_0},r)}$. This completes the proof since $z_0\mapsto\bar{z}_0$ is a bijection $U\rightarrow V$.
