Prove a function is holomorphic 
Let 
  
  
*
  
*$A \subset \mathbb{C}$ be an open set
  
*$f : A \rightarrow \mathbb{C}$ be a holomorphic function
  
*$B = \{z \in \mathbb{C} | \bar{z} \in A\}$
  
*$g(z) : B \rightarrow \mathbb{C} = \overline{f(\bar{z})}$
  
  
  Show by using the Cauchy-Riemann equations that $g$ is holomorphic for
  $B$.

I do have problems to understand the Cauchy Riemann equations. As far as I understand I need to show that g is complex differentiable $\forall z \in B$, and therefore the first derivate at $g(z)$ exists, right? The Cauchy-Riemann equations allow to swap some arguments of partial derivatives, but how can this be used here?
 A: Write
$$f(x,y)=u(x,y)+i v(x,y),\quad g(\xi,\eta)=a(\xi,\eta)+ib(\xi,\eta)$$
where $u$, $v$, $a$, $b$ are realvalued functions defined in $A$, resp. $B$. Then by definition of $g$ one has
$$a(\xi,\eta)+ib(\xi,\eta)=g(\xi+i\eta)=\overline{f(\xi-i\eta)}=u(\xi,-\eta)-iv(\xi,-\eta)$$
and therefore
$$a(\xi,\eta)=u(\xi,-\eta),\quad b(\xi,\eta)=-v(\xi,-\eta)\qquad\bigl((\xi,\eta)\in B\bigr)\ .$$
It follows that, e.g. $$b_\eta(\xi,\eta)=-v_y(\xi,-\eta)\cdot(-1)=v_y(\xi,-\eta)\ .$$
Since $u$ and $v$ satisfy the CR-equations in the variables $x$ and $y$ we conclude  that
$$a_\xi(\xi,\eta)=u_x(\xi,-\eta)=v_y(\xi,-\eta)=b_\eta(\xi,\eta)\ ,$$
and similarly
$$a_\eta(\xi,\eta)=-u_y(\xi,-\eta)=v_x(\xi,-\eta)=-b_\xi(\xi,\eta)\ .$$
This shows that $g$ fulfills the CR-equations in the variables $\xi$ and $\eta$.
But there is also a  direct approach, which in my view is simpler and more in tune with a complex world description.
As $f$ is holomorphic in $A$, for each point $z_0\in A$ (held fixed in the following) there is a complex number $C$ such that
$$f(z)-f(z_0)=C(z-z_0)+o(|z-z_0|)\qquad (z\in A, \ z\to z_0)\ .$$
Let a point $w_0\in B$ be given, and put $z_0:=\bar w_0$. Then by definition of $g$ one has
$$g(w)-g(w_0)=\overline{f(\bar w)}-\overline{f(\bar w_0)}=\overline{f(\bar w)-f( z_0)}=\overline{C(\bar w -z_0)+o(|\bar w-z_0|)}\qquad(w\in B)\ .$$
As $|\bar w -z_0|=|w-w_0|$ it follows that
$$g(w)-g(w_0)=\bar C(w-w_0)+o(|w-w_0|)\qquad(w\in B, \ w\to w_0)\ .$$
It follows that $g'(w_0)=\bar C$, and as $w_0\in B$ was arbitrary, we conclude that $g$ is holomorphic in $B$.
A: You can also prove it using the definition. Everything works nicely since the conjugation $z\mapsto\overline{z}$ is continuous. 
For any $z_0$ in $B$, we have $\frac{g(z)-g(z_0)}{z-z_0}=\frac{\overline{f(\overline{z}})-\overline{f(\overline{z_0})}}{z-z_0}=\overline{\ \frac{f(\overline{z})-f(\overline{z_0})}{\overline{z}-\overline{z_0}}\ }$.
But, by continuity of the conjugation, $z\to z_0$ if and only if $\overline{z}\to\overline{z_0}$, so 
$\displaystyle{ 
\lim_{z\to z_0} \frac{g(z)-g(z_0)}{z-z_0}
=
\overline{ 
\lim_{\overline{z}\to\overline{z_0}}
\frac{ f(\overline{z})-f(\overline{z_0}) }{ \overline{z}-\overline{z_0}}
}
=\overline{f'(\overline{z_0})} 
}$
(the last equality justified by the fact that $\overline{z_0}$ belongs to $A$).
