Answer
Using both rectangular and polar expressions for $\Delta f$, we show that $\Delta f = 0$. Hence, $f\left( {x,y} \right) = {\tan ^{ - 1}}\frac{y}{x}$ is harmonic.
Work Step by Step
We have $f\left( {x,y} \right) = {\tan ^{ - 1}}\frac{y}{x}$.
We evaluate the Laplace operator using
1. rectangular expression
Referring to Theorem 2 of Section 7.8, the partial derivatives are
${f_x} = \frac{1}{{{y^2}/{x^2} + 1}}\left( { - \frac{y}{{{x^2}}}} \right) = - \frac{y}{{{x^2} + {y^2}}}$, ${\ \ }$ ${f_y} = \frac{1}{{{y^2}/{x^2} + 1}}\left( {\frac{1}{x}} \right) = \frac{x}{{{x^2} + {y^2}}}$
${f_{xx}} = \frac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$, ${\ \ }$ ${f_{yy}} = - \frac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
Therefore, $\Delta f = {f_{xx}} + {f_{yy}} = 0$.
Since $\Delta f = 0$, $f$ is harmonic.
2. polar expression
In polar coordinates, $x = r\cos \theta $ and $y = r\sin \theta $. So, $f$ becomes
$f\left( {r,\theta } \right) = {\tan ^{ - 1}}\frac{{r\sin \theta }}{{r\cos \theta }} = \theta $
The partial derivatives are
${f_r} = 0$, ${\ \ \ }$ ${f_\theta } = 1$
${f_{rr}} = 0$, ${\ \ \ }$ ${f_{\theta \theta }} = 0$
Using Eq. (13) we get
$\Delta f = {f_{rr}} + \frac{1}{{{r^2}}}{f_{\theta \theta }} + \frac{1}{r}{f_r}$
$\Delta f = 0$
Since $\Delta f = 0$, $f$ is harmonic.
