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.

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    $\begingroup$ Try $a=b=1$ and see what you get in the hint $\endgroup$ – Conrad Sep 26 at 18:22
  • $\begingroup$ Thank you. I'm curious: how did you determine a and b? $\endgroup$ – Tom Sep 26 at 18:41
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    $\begingroup$ 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 $1$) $\endgroup$ – 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.


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