Why is $\sqrt r \cos \frac \theta 2$ harmonic?

Question

Why is $\sqrt r \cos \frac \theta 2$ harmonic?

Answer 1

$$\sqrt{r}\cos \frac{\theta}{2}=Re(\sqrt{re^{i\theta}})=Re(\sqrt z)$$ The components of analytical functions are harmonic, so this is it.

Answer 2

Keeping in mind that, in polar coordinates, the Laplacian looks like

$$\Delta = \frac {\partial^2} {\partial r^2} + \frac 1 r \frac \partial {\partial r} + \frac 1 {r^2} \frac {\partial^2} {\partial \theta^2}$$

and that "harmonic" means $\Delta f = 0$, it easy to check that in your case

$$\Delta f = - \frac 1 4 r^{-\frac 3 2} \cos \frac \theta 2 + \frac 1 r \left( \frac 1 2 r^{-\frac 1 2} \cos \frac \theta 2 \right) + \frac 1 {r^2} \left( -\frac 1 4 r^{\frac 1 2} \cos \frac \theta 2 \right) = 0 .$$

Unlike the approach based upon finding an analytic function such that $f$ should be its real or imaginary part, the approach based upon computing the Laplacian has the advantage of not depending on "inspiration", being purely computational (and thus algorithmic).