Proving a function is harmonic without calculating partial derivatives

Question

This is for an assignment, so I don't want any explicit solutions, but just some kind of idea how to continue.

I'm trying to prove that the family of functions $u_{\xi}(z)=\frac{1-|z|^{2}}{|\xi-z|^{2}}$ are harmonic in the complex unit ball, with $\xi$ being a complex constant so $|\xi|=1$. Normally my method would be trying to look at the second partial derivatives, but initial analysis with Maple makes them look highly unpleasant. In the assignment, I am hinted that I should construct a holomorphic function with real part $u_{\xi}$, which I know I can do by finding the primitive of $\frac{\delta u}{\delta x}- i\frac{\delta u}{\delta y}$. However, that expression also seems very unpleasant to work with, let alone integrate. Is there some other method I could use?

Answer 1

First note that since $$ u_{\xi}(z)=\frac{1-|z|^{2}}{|\xi-z|^{2}} = \frac{1-|z/\xi|^{2}}{|1-z/\xi|^{2}} $$ it suffices to consider the case $\xi = 1$, i.e. to show that $$ u(z) = \frac{1-|z|^{2}}{|1-z|^{2}} $$ is harmonic in the unit disk. This makes the the calculation of $\frac{\partial u}{\partial x}- i\frac{\partial u}{\partial y}$ manageable, you'll ultimately get

$$ \frac{2((x-1)^2 - y^2) - 4i(x-1)y}{((x-1)^2 + y^2)^2} = \frac{2(x - 1 - iy)^2}{((x-1)^2 + y^2)^2} $$

which can be identified as the holomorphic function

$$ \frac{2 (\bar z - 1)^2}{|z-1|^4} = \frac{2}{(1-z)^2} $$

An alternative is to rewrite $u$ in a way that it can be "recognized" as the real part of a holomorphic function (which admittedly is easier if one knows the solution already). In your case, $$ \frac{1-|z|^{2}}{|1-z|^{2}} + C = \frac{1-z \bar z}{1 - z - \bar z + z \bar z} + C $$ with a suitable real constant $C$ does the trick (try to choose $C$ such that the numerator simplifies).