For $(x,y) \in D(0,1)$ fixed, we define the complex number $z = x + iy$ and we set for $t$ real:
$$\mathrm{N}(x,y,t) = \frac{1 - |z|^2}{|z - e^{it}|^2} = \frac{1 - (x^2 + y^2)}{(x - \cos t)^2 + (y - \sin t)^2}$$
Deduce that $\dfrac{1}{2\pi} \int_0^{2\pi} \mathrm{N}(x,y,t)\, \mathrm{d}t = 1$.
One may write $\dfrac{1}{1 - ze^{-it}}$ in the form of the sum of a series of functions.