\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

Defining $\log(z-a)=\log|z-1|+i\arg(z-a)$, show that the real and
imaginary parts of $\log(z-a)$ are harmonic functions in any
region not containing $z=a$.

\medskip

{\bf Answer}

\un{Method 1}

If $R$ does not contain $a$, then $w=\ln(z-a)$ is analytic in $R$.
Hence the real and imaginary parts are harmonic in $R$.

\un{Method 2}

Let $z-a=re^{i\theta}$. Then $\log(z-a)=\log r+i\theta$.

Then in polars, Laplace's equation is

$$\ds\frac{\pl^2\phi}{\pl r^2}+\ds\frac{1}{r}\ds\frac{\pl
\phi}{\pl r}+\ds\frac{1}{r^2}\ds\frac{\pl^2\phi}{\pl \theta^2}=0$$

Can substitute $\phi=\log r$ \un{or} $\phi=\theta$ and they
automatically satisfy this equation. Hence $Re$ and $Im$ parts are
harmonic.

\end{document}
