$f(z)$ and $\overline{f(\overline{z})}$ simultaneously holomorphic 
Prove that the functions $f(z)$ and $\overline{f(\overline{z})}$ are simultaneously holomorphic.

I take this to mean that $f(z)$ is holomorphic if and only if $\overline{f(\overline{z})}$ is holomorphic.
Let $g(z)=\overline{f(\overline{z})}$. Note that $\overline{g(\overline{z})}=f(z)$. So it suffices to prove that if $f(z)$ is holomorphic, then $g(z)$ is holomorphic.
Write $f(z)=u(z)+iv(z)$. Since $f(z)$ is holomorphic, the real and imaginary parts satisfy the Cauchy-Riemann equations:
$$\frac{\partial{u(z)}}{\partial{x}} = \frac{\partial{v(z)}}{\partial{y}}, \frac{\partial{u(z)}}{\partial{y}} = -\frac{\partial{v(z)}}{\partial{x}}.$$
We have $g(z) = u(\overline{z})+i(-v(\overline{z}))$. To prove that $g(z)$ is holomorphic, we must prove that its real and imaginary parts satisfy the Cauchy-Riemann equations:
$$\frac{\partial{u(\overline{z})}}{\partial{x}} = \frac{\partial{(-v(\overline{z}))}}{\partial{y}}, \frac{\partial{u(\overline{z})}}{\partial{y}} = -\frac{\partial{(-v(\overline{z}))}}{\partial{x}}.$$
How can we obtain this from the above relations?
 A: Proof based in holomorphic $\implies$ locally power series:
$$f(z) = \sum_{n=0}^\infty a_n (z-c)^n,$$
$$
\tilde f(z) = \overline{f(\overline{z})} =
\overline{\sum_{n=0}^\infty a_n (\overline{z}-c)^n} =
\sum_{n=0}^\infty\overline{a_n (\overline{z}-c)^n} =
\sum_{n=0}^\infty\overline{a}_n(z-\overline{c})^n,
$$
where the third equality is true because the continuity of conjugation.
A: I know that I am not answering your final question, but anyway... if $g(z)=\overline{f(\overline z)}$, then
$$\begin{align*}
g^\prime(a)=&\,\lim_{z\to a}\frac{g(z)-g(a)}{z-a}\\[2mm]
=&\,\lim_{z\to a}\frac{\overline{f(\overline z)}-\overline{f(\overline a)}}{z-a}\\[2mm]
=&\,\lim_{z\to a}\frac{\overline{f(\overline z)-f(\overline a)}}{\overline{\ \overline{z-a}\ }}\\[2mm]
=&\,\lim_{z\to a}\overline{\,\Biggl[\frac{f(\overline z)-f(\overline a)}{\overline z-\overline a}\Biggr]}\\[2mm]
=&\,\overline{\lim_{z\to a}\,\frac{f(\overline z)-f(\overline a)}{\overline z-\overline a}}\\[2mm]
=&\,\overline{\lim_{w\to\overline a}\,\frac{f(w)-f(\overline a)}{w-\overline a}}\\[2mm]
=&\overline{f^\prime(\overline a)}\,.
\end{align*}$$
Thus, $g$ is holomorphic. The converse is proved similarly (or you can use the fact that the transformation $f\mapsto g$ is idempotent, that is, you return to your original function $f$ when applied twice).
A: Morera's theorem is an extremely useful tool when one wants to prove a function is holomophic but doesn't want to look at derivatives explicitly. In this case, let $\gamma$ be a simple closed curve.
$\int _\gamma \overline{f(\overline{z})} dz = \overline{\int _\gamma f(\overline{z}) dz} = \int _\gamma f(\overline{z}) d\overline{z} = -\int_{\overline{\gamma}} f(u) du$
where $\overline{\gamma}$ is the curve $\gamma$, reflected with respect to the x-axis and the - comes from the corresponding change of the orientation from counterclockwise to clockwise.
Now, by the above inequality and Morera's theorem:
$f(z)$ is holomorphic iff its integral over an arbitrary simple closed curve is $0$ iff $\overline{f(\overline{z})}$ is holomorphic, which we wanted to prove.
