Constant function Let $z \mapsto f(z)$ and $z \mapsto \overline{f(z)}$ be analytic functions on a plane $\Omega$. Show that $f$ is constant function.
I know that $f$ is constant if it's derivate is $0$ on all $\Omega$? And let $z$ be $x+iy$. 
$f(z)=u(x,y)+iv(x,y)$ and $\overline{f(z)}=u(x,y)-iv(x,y)$. 
If $z$ maps to $f(z)$ and $\overline{f(z)}$ does it mean $f(z)=\overline{f(z)}$? so $v(x,y)=-v(x,y)$. If $f'(z)=0 \in \Omega$ does it mean the CR equations hold only when $u_x=u_y=v_x=v_y=0$?
 A: It is not necessarily the case that $f(z) = \overline{f(z)}$. A priori, $z \mapsto f(z)$ and $z \mapsto \overline{f(z)}$ could be two different analytic functions.
Having said that, you are very close. 
As $z \mapsto f(z)$ is analytic, $u(x, y) = \operatorname{Re}(f(x+iy))$ and $v(x, y) = \operatorname{Im}(f(x+iy))$ satisfy the Cauchy-Riemann equations, so $u_x = v_y$ and $u_y = -v_x$. 
As $z \mapsto \overline{f(z)}$ is analytic, $u(x, y) = \operatorname{Re}\left(\overline{f(x+iy)}\right)$ and $-v(x, y) = \operatorname{Im}\left(\overline{f(x+iy)}\right)$ satisfy the Cauchy-Riemann equations, so $u_x = (-v)_y = -v_y$ and $u_y = -(-v)_x = v_x$.
Can you proceed from here to show that $u$ and $v$, and hence $f$, are constant?
A: Theorem: if $f$ is a non constant analytic function, then $f$ is open.
If $f$ and $\overline{f}$ are analytics then $g(z)=f(z)\overline{f(z)}=|f(z)|^2\in \Bbb R$  is also analytic hence it is constant $r\geq 0$ (because $\Bbb R$ is not open in $\Bbb C$), in particular $f$  lie in $S(0,r)$(=the circle of center $0$ and radius $r$ ) which is not open in $\Bbb C$, it follow that $f$ is constant. 
A: The Cauchy-Riemann equations for $f$ are
$$u_x = v_y \\
u_y = -v_x$$
The Cauchy-Riemann equations for $\overline{f}$ are 
$$u_x = -v_y \\
u_y = v_x$$
Now try adding some of these together and see what happens.
