Show that the function $u(z)=\frac{1-|z|^2}{|1-z|^2}$ is harmonic in $K(0,1)$: the open disc with center 0 and radius 1, and the determine the conjugate harmonic functions to $u$.
I've been given the following hint:
Determine $a,b \in \mathbb{C}$ such that $\Re(\frac{az+b}{1-z})=u(z)$.
However, I don't know how to go about it.
In order to check whether a function is harmonic, I usually just check the condition $\partial^2/\partial x^2+\partial^2/\partial y^2=0$, but that doesn't seem appropriate here, hence the hint.
Best Answer
One way to go about it is to develop the expression as $$\frac{az+b}{1-z} = \frac{(az+b)(1-\bar z)}{|1-z|^2} = \frac{az + b - b \bar z - a|z|^2}{|1-z|^2}.$$ This highly suggests taking $a=1$ and $b=1$ because then you get $1 - |z|^2$ in the numerator. The remaining term is $z - \bar z$ which is imaginary.