My attempt to show that $u(r e^{i \theta}) = \theta \log r$ is harmonic is the next: if $z = r e^{i \theta}$ we have $\log z = \log r + i \theta$, then ${(\log z)}^2 = {(\log r)}^2 + i 2 \theta \log r - {\theta}^2 = {(\log r)}^2 + i 2 u(z) - {\theta}^2$; that means that $u(z) = \frac{Im [(\log z)^2]}2 = Im \varphi$, where $$ \varphi(z) = {\left(\frac{\log z}{\sqrt{2}}\right)}^2 $$ As $\varphi$ is holomorphic and $u$ is the imaginary part of $\varphi$, then $u$ is harmonic.
Do you think that my argument is correct?