Harmonic function in the upper half plane 
Find an harmonic function $h(z)$ in the upper half plane with the
  following properties:
  
  
*
  
*$h$ is bounded in $\mathbb{C}_+$ and is continuous in $\{z | \Im(z) \geq 0 \}\setminus \{0 \}$
  
*$h(x)=1$ if $x>0$ and $h(x)=-1$ if $x<0$
Also, prove that there is exactly one harmonic function with these
  properties.

This time I really  have no clue. Any hints?
 A: If we consider $f(z)=f(x,y)$, then this problem is equivalent to the 1st boundary problem of Laplas' equation:
$$\begin{cases}\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial x^2}=0 \\
|f(x,y)|\le C \text{   for all } x \in \mathbb{R} \text{ and for } y>0\\
f(x,0)=\begin{cases}1,x>0\\
-1,x<0
\end{cases}
\end{cases}$$
It has very simple solution with the help of fundamental operator:
$$f(x,y)=\frac{y}{\pi}\int\limits_{-\infty}^{+\infty}\frac{f_{0}(t)}{\left(x-t\right)^2+y^2}dt= -\frac{y}{\pi}\int\limits_{-\infty}^{0}\frac{1}{\left(x-t\right)^2+y^2}dt+\frac{y}{\pi}\int\limits_{0}^{+\infty}\frac{1}{\left(x-t\right)^2+y^2}dt$$
If you want to represent this function as a complex value function, you can calculate this integrals. The result is
$$f(x,y)=\operatorname{csgn}(\bar{y})\frac{\pi}{y} $$
-- this is that stuff which the program has calculated. In complex it is:
$$ f(z)=\operatorname{csgn}(\overline{\operatorname{Im} z})\frac{\pi}{\operatorname{Im} z}$$
According to the uniqueness of the solution, this solution is unique.
