$u(z)$ harmonic if and only if $u(\overline{z})$ harmonic 
Prove that the function $u(z)$ is harmonic if and only if $u(\overline{z})$ is harmonic.

$u(z)$ harmonic means that $$\dfrac{\partial^2 u(x,y)}{\partial x^2} + \dfrac{\partial^2 u(x,y)}{\partial y^2} = 0.$$We must prove that $$\dfrac{\partial^2 u(x,-y)}{\partial x^2} + \dfrac{\partial^2 u(x,-y)}{\partial y^2} = 0.$$
I think the chain rule should be used to prove this, but I'm not so familiar with how to write it down here. Any help?
 A: Talking about the function "$\ u(\bar z)\ $" means defining a new function
$$g(x,y):=u(x,-y)\ .$$
Then obviously $g_{.11}(x,y)=u_{.11}(x,-y)$, and according to the chain rule
$$g_{.2}(x,y)={\partial\over\partial y}\bigl(u(x,-y)\bigr)=u_{.2}(x,-y)\cdot(-1)\ ,$$  $$g_{.22}(x,y)={\partial\over\partial y}\bigl(u_{.2}(x,-y)\cdot(-1)\bigr)=
 u_{.22}(x,-y)\cdot (-1)^2\ .$$
(The notation ${}_{.k}$ means differentiation with respect to the $k^{\rm th}$ variable of the outer function.) It follows that
$$\Delta g(x,y)=\Delta u(x,-y)\quad\forall (x,y)\ .$$
A: Let $u(z)$ harmonic function in open set $D\subset\mathbb{C}$, it's clear that $u(\bar{z})$ can not be harmonic in $D$. For example $u(z)=\frac{1}{z+i}$ is harmonic in $\mathbb{C}\setminus\{-i\}$ and $u(\bar{z})$ is harmonic in $\mathbb{C}\setminus\{i\}$. So if $u(z)$ is harmonic in $D$ we define the trasformation $F:D\subset \mathbb{C} \longrightarrow \mathbb{C}$ such that $F(z)=\bar{z}$ and $u(\bar{z})$ is harmonic in $\bar{D}=F(D)$. Infact we can compute now for $z\in \bar{D}$ 
$$\Delta u(\bar{z})=\Delta u(\zeta)=0$$
for some $\zeta\in D$ because $F(F(z))=z$.
