# Show that $u(z)=\frac{1-|z|^2}{|1-z|^2}$ is harmonic in the open disc with center 0 and radius 1

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.

• Try $a=b=1$ and see what you get in the hint – Conrad Sep 26 at 18:22
• Thank you. I'm curious: how did you determine a and b? – Tom Sep 26 at 18:41
• see the answer below - also $u(z)$ above is a well known positive harmonic function and the prototypical one generated by integrating the Poisson kernel against a singular positive measure on the unit circle (the Dirac measure at $1$) – Conrad Sep 26 at 19:32

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.