I'm trying to show that given $u:\bar D(0,R)\subset \Omega\rightarrow\mathbb R$ an harmonic function, one has that: $$\forall z\in D(0,R): u(z)=\operatorname{Re}\left(\frac{1}{2\pi i}\oint_{\partial D(0,R)}\frac{\zeta-z}{\zeta(\zeta-z)}u(\zeta)\,d\zeta\right)\ [\star]$$
Thanks in advance.
My try:
- Firstly, I've tried expressing the harmonic function using the mean valued proprety derived from the Cauchy formula for $f$: $$u(0)=\frac{1}{2\pi}\int_0^{2\pi}u(z+Re^{it})dt$$ and extending on a collection of smaller disks but that didn't work,
- Secondly I',ve tried to start from th Cauchy integral formula for $f$ and express $u$ as the real part but I can't seem to obtain the term $\frac{\zeta-z}{\zeta}$ inside the integral$\ [\star]$
- I also tried some calculations with Cauchy-Riemann but it didn't lead anywhere.
NB: I would like to do without the poisson formula, if it is possible.