= Solution
For a radius $\rho>0$, write $z=\rho e^{i\theta}$. Then
$$
w=(\rho+\rho^{-1})\cos\theta+i(\rho-\rho^{-1})\sin\theta.
$$
For $\rho\ne1$ its image is the <ellipse>
$$
\boxed{\frac{(\operatorname{Re}w)^2}{(\rho+\rho^{-1})^2}
+\frac{(\operatorname{Im}w)^2}{(\rho-\rho^{-1})^2}=1.}
$$
Its horizontal semiaxis is $\rho+\rho^{-1}$, its vertical semiaxis is $|\rho-\rho^{-1}|$, and its <foci> are $\pm2$ because the squared semiaxes differ by $4$. Reciprocal radii give the same <ellipse>, traversed in opposite directions as $\theta$ increases.
\Image[/past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012-s1-joukowski-circles.png]
{title=The Joukowski map sends reciprocal-radius circles to one ellipse and the unit circle to a doubled interval}
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