TITLE: Analyticity of $\overline {f(\bar z)}$ given $f(z)$ is analytic
QUESTION [10 upvotes]: Suppose $f$ is an analytic function on a domain $D$. Then I need to show that $\overline {f(\bar z)}$ is also analytic. 
Here is what I did - 
Suppose $f(z) = u(x,y) + iv(x,y)$ where $u$ and $v$ are real functions of $x$ and $y$ and $z = x + iy$. Now $f(\bar z) =  u(x,-y) + iv(x,-y) $ and then $\overline {f(\bar z)} = u(x,-y) - iv(x,-y) $. 
To show that a function is analytic, I need to verify that it satisfy Cauchy-Riemann Equations. I differentiated $\overline {f(\bar z)}$ and checked that it actually satisfies these equations. But here is my doubt - Being analytic means that the function if complex differentiable. Now here to check for analyticity, I am differentiating my function without proving that it's actually analytic. So is this the right way to do it?

REPLY [4 votes]: Since $f$ is analytic, it can be represented as a power series, that is, for every $z_o\in D$,
$$f(z)=\sum_{n=0}^\infty c_n(z-z_0)^n$$
Let's assume that $z_0=0$. Then
$$\overline {f(\overline z)}=\sum_{n=0}^\infty c_n\overline {{z}}^n=\overline{ \lim_{k\to \infty}\sum_{n=0}^k c_n\overline{z}^n}$$
Since the function $z\mapsto \overline z $ is continuous,
$$\overline{ \lim_{k\to \infty}\sum_{n=0}^k c_n\overline{z}^n}=\lim_{k\to \infty}\overline{\sum_{n=0}^kc_n\overline{z}^n}= \lim_{k\to \infty}\sum_{n=0}^k \overline{c_n\overline{{z}^n}}=\lim_{k\to \infty}\sum_{n=0}^k\overline{c_n}z^n=\sum_{n=0}^{\infty}\overline{c_n}z^n$$
But
$$|c_nz^n|=|c_n||z^n|=|\overline{c_n}||z^n|=|\overline{c_n}z^n|$$
and then $\sum_{n=0}^{\infty}c_nz^n$ is absolutely convergent if and only if $\sum_{n=0}^\infty\overline{c_n}z^n$ is absolutely convergent. But $f(z)$ is analytic in $D$, so $\sum_{n=0}^{\infty}c_nz^n$ is absolutely convergent; then we can conclude that
$$\overline {f(\overline z)}=\sum_{n=0}^\infty \overline{c_n}{z}^n$$
is also analytic in $D$.