= Solution
The <Einstein angular radius> is the image-ring radius of an exactly aligned source. For the <singular isothermal sphere lens>, set $\beta=0$ in <5/c/ii>:
$$
\boxed{\theta_E=4\pi\left(\frac{\sigma_{\mathrm{DM}}}{c}\right)^2\frac{D_{LS}}{D_S},\qquad
\beta=\theta-\theta_E\operatorname{sgn}\theta.}
$$
Choose $\beta>0$ without loss of generality. The positive branch always gives $\theta_+=\beta+\theta_E$. A negative-side image would have $\theta_-=\beta-\theta_E$; this branch is admissible only when its result is negative. Therefore
$$
\boxed{\begin{array}{c|l}
0<\beta<\theta_E&\theta_+=\beta+\theta_E,\quad\theta_-=\beta-\theta_E\quad\text{two images},\\
\beta>\theta_E&\theta_+=\beta+\theta_E\quad\text{one image}.
\end{array}}
$$
At $\beta=0$, symmetry produces an <Einstein ring> of angular radius $\theta_E$. At $\beta=\theta_E$, the negative branch reaches the singular origin and is not an ordinary additional finite image; the one finite image is at $2\theta_E$. Reversing the source sign reverses the signed image coordinates, so the multiplicity criterion uses $|\beta|$. These features rely on the singular isothermal profile; a smooth finite-density core changes the central-image behaviour.
\Image[/past-exam-of-the-astrophysics-course-of-the-university-of-oxford/2018-isothermal-lens-mapping.png]
{title=Two branches of the singular isothermal lens equation}
{description=The normalized map is $\beta/\theta_E=\theta/\theta_E-\operatorname{sgn}\theta$. Horizontal source lines show two images inside the Einstein radius and one outside. The open endpoints exclude the singular lens centre.}
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