Let $z\in\Omega \subset\mathbb{C}$ and $z_p\notin \Omega$. Show that $\text{Im}(\frac{z^2}{z-z_p})$ is harmonic in $\Omega$, where $\text{Im}(z)$ is the imaginary part of $z$.
So far: For $z = \alpha + i\beta$, $z_p = \xi + i\eta$, I have $\text{Im}(\frac{z^2}{z-z_p}) = -\frac{(\alpha^2 -\beta^2)(\beta - \eta) - 2\alpha\beta(\alpha - \xi)}{(\alpha - \xi)^2 + (\beta - \eta)^2}$, but would rather not differentiate it under the Laplace operator.
Question: Is there any easier, that is less technical, way of showing that $\text{Im}(\frac{z^2}{z-z_p})$ is harmonic in $\Omega$?
$f(z) = \dfrac{z^2}{z-z_p}$ is holomorphic (except at $z=z_p$). It follows from Cauchy-Riemann's equations that the real and imaginary parts of a holomorphic function are harmonic.