Derive Poisson's integral formula for Im z>0 How to derive Poisson's integral formula for $\text{Im }{z}>0$ given that for $|z|<1 $ we have $$f(re^{i\theta})=\frac{1}{2\pi}\int_0^{2\pi}\frac{1-r^2}{1+r^2-2r\cos(\theta-\phi)}f(e^{i\phi})d\phi$$
 A: Here's a way to do it by using only the simplest instance of Poisson's formula on the disk, namely $f(0)=\frac{1}{2\pi}\int_0^{2\pi} f(e^{i\phi})\,d\phi$. 
We have some boundary data $g:\mathbb R\to\mathbb R$ and wish to find a harmonic function $U$ on the upper half-plane $\mathbb H$ for this data. Fix a point $x_0+iy_0\in \mathbb H$ and consider the conformal map $F(z)=x_0+iy_0\frac{1+z}{1-z}$. Note that $F$ maps the unit disk onto $\mathbb H$ and $F(0)=x_0+iy_0$. Pull back the boundary data: $g\circ F$  is defined on the unit circle. Let $u$ be the harmonic extension of $g\circ F$ to the disk. By the above, 
$$u(0)=\frac{1}{2\pi}\int_0^{2\pi} (g\circ F)(e^{i\phi})\,d\phi\tag1$$
On the other hand, $u=U\circ F$ by conformal invariance of harmonic functions. Therefore, the integral in (1) actually gives us $U(x_0+iy_0)$, which is what we wanted. It remains to change the variable of integration in (1), so that integral is taken over $\mathbb R$, in the original domain of definition of $g$. This yields
$$\frac{1}{2\pi}\int_0^{2\pi} (g\circ F)(e^{i\phi})\,d\phi 
= \frac{1}{2\pi}\int_{\mathbb R} g(x) |G'(x)|\,dx \tag2$$
where $G=F^{-1}$. I leave the computation of $G'$ to you.
