= Solution
The principal real arctangent is defined here on $x\ne0$, so work on either of the two open half-planes. Put $r=\sqrt{x^2+y^2}$. Direct <differentiation> gives
$$
\phi_x=-\frac{y}{r^2},\qquad\phi_y=\frac{x}{r^2},\qquad\phi_{xx}=\frac{2xy}{r^4},\qquad\phi_{yy}=-\frac{2xy}{r^4}.
$$
Hence $\phi_{xx}+\phi_{yy}=0$: \b[$\phi$ is harmonic on its domain]. The <Cauchy-Riemann equations> for $f=\phi+i\psi$ require
$$
\psi_x=-\phi_y=-\frac{x}{r^2},\qquad\psi_y=\phi_x=-\frac{y}{r^2}.
$$
Integrating these <partial derivatives> gives the <harmonic conjugate>
$$
\boxed{\psi(x,y)=-\log r+C_0=-\frac12\log(x^2+y^2)+C_0.}
$$
The constants may be chosen independently on the two components. The <harmonic angle and logarithmic radius> representation supplies the <holomorphic function>
$$
\boxed{f(z)=\begin{cases}-i\operatorname{Log}z+iC_0,&\operatorname{Re}z>0,\\-i\operatorname{Log}(-z)+iC_0,&\operatorname{Re}z<0,\end{cases}}
$$
where $\operatorname{Log}$ is the <branch of the complex logarithm> on the right half-plane. On the left half-plane it is the angle of $-z$, rather than the principal angle of $z$, that equals $\arctan(y/x)$. This distinguishes the literal arctangent from a globally defined argument; no single-valued <complex logarithm> exists on the whole punctured plane.
With $C_0=0$, the <level sets> $\phi=C$, for $-\pi/2<C<\pi/2$, are the two opposite rays of $y=x\tan C$ with the origin removed. On each half-plane only one ray remains. The <level sets> $\psi=K$ are circles $r=e^{-K}$, with their two points on $x=0$ excluded from the literal domain. The rays and circles meet at right angles, as expected for <harmonic conjugates>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-3-harmonic-contours.png]
{title=Angle contours are opposite rays; logarithmic-radius contours are circles, on the two half-planes $x\ne0$.}
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