Is it true or not : if $u(z)$ is harmonic, $u(\overline{z})$ is also harmonic. Is it true or not :  if $u(z)$ is harmonic, $u(\overline{z})$ is also harmonic.
My try :
$u(z)=u(x,y)$ is harmonic
Define $s=-y$
Let $U := u(\overline{z})=u(x,-y)=u(x,s)$ : $$\frac{\partial U}{\partial x}=\frac{\partial u}{\partial x} \Rightarrow \frac{\partial^2 U}{\partial x^2}=\frac{\partial^2 u}{\partial x^2}$$
And $$\frac{\partial U}{\partial y}=\frac{\partial u}{\partial x} \frac{\partial x}{\partial y} + \frac{\partial u}{\partial s} \frac{\partial s}{\partial y}  = - \frac{\partial u}{\partial s}  $$
Similarly $$\frac{\partial^2 U}{\partial y^2}=   \left[ \frac{\partial }{\partial y} ( \frac{\partial U}{\partial y}) \right] =- \left[ \frac{\partial }{\partial y} ( \frac{\partial u}{\partial s}) \right] = ... = - \left[ - \frac{\partial }{\partial s} ( \frac{\partial u}{\partial s}) \right] = \frac{\partial^2 u}{\partial s^2}  $$ Hence
$$\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial y^2} = \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial s^2}= 0 \ \ \ (*)$$
Therefore $u(\overline{z})$ is also harmonic.
*My Questions is : Is my try problematic?Does $(*)$ needs justification? *

A funciton $\phi$is harmonic if 
$$ \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0$$
It can be shown that $$ \frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} = \frac{\partial}{\partial z}\frac{\partial}{\partial \bar{z}}$$
This means for a function of a complex variable to be harmonic it needs to be an analytic function of just $z$ or an analytic function of just $\bar{z}$. Analytic so that the derivative exists and a function of just one or the other so that the result is zero. 
This means that the real part and the imaginary part of said analytic function will each be harmonic functions 

The following is true only if $f(z)$ is real when $z$ is real
Note that many cases nothing new is gained in making it a function of $\bar{z}$ since for certain analytic functions this is equivalent to taking the complex conjugate.
$$ f(\bar{z}) = \overline{f(z)} = \overline{u(x,y)+iv(x,y)} = u(x,y)-iv(x,y) $$
So that we don't really get fundamentally new solutions to Laplace's equation.

To obtain the first equality note that when a function is analytic within some neighborhood of a point there is a Taylor series which converges to the function in within some open set containing that point. This allows us to write,
$$f(z) =  \sum_{n=0}^N a_n z^n + R_N(z)$$
Where $R_N \rightarrow 0$ as $N \rightarrow \infty$
$$ \overline{ f(z) } = \overline{\sum_{n=0}^N a_n z^n + R_N} = \overline{\sum_{n=0}^N a_n z^n} + \overline{R_N} = \sum_{n=0}^N \overline{a_n} (\overline{z})^n + \overline{R_N}$$
Since $\vert \overline{R_N} \vert = \vert R_N \vert$ we can conclude that $\overline{R_N} \rightarrow 0 $ whenever $R_N$ does. If the coefficientes $a_n$ are real then we can write $\overline{a_n} = a_n$ and conclude that,
$$ \overline{f(z)} = \sum_{n=0}^\infty a_n (\overline{z})^n = f(\bar{z})$$
