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 / 2

The Joukowski map $f(z)=z+z^{-1}$ is holomorphic on $\mathbb{C}\setminus \{0\}$. A holomorphic function is locally conformal where its derivative is nonzero. Here $f^{\prime }(z)=1-z^{-2}$ vanishes at $z=\pm 1$. Thereforewhere the exclusion of zero is because the function is undefined there. Holomorphy alone does not supply conformality at the two critical points.

Put $z=1+\zeta$. Expanding the holomorphic function near its critical point givesFor a small ray $\zeta =r{e}^{i\theta }$, the leading image displacement is $r^2{e}^{2i\theta }$. Thus oriented ray angles are doubled modulo $2\pi$ at $z=1$. This is angle multiplication at a holomorphic critical point of multiplicity two; an unoriented angle between full tangent lines must also be interpreted modulo $\pi$. The map is not locally one-to-one there.

For $z=e^{i\theta }$ the image is $w=2\cos \theta$, a real number. Therefore each semicircle maps onto the real interval $[-2,2]$, with endpoints included for closed semicircles. Under positive traversal of the full unit circle, the upper half runs from $2$ to $-2$ and the lower half returns from $-2$ to $2$. Every interior point has two unit-circle preimages. The Joukowski circle-to-ellipse mapping degenerates to a doubled line segment at this radius.

For a radius $\rho >0$, write $z=\rho e^{i\theta }$. ThenFor $\rho \ne 1$ its image is the ellipseIts horizontal semiaxis is $\rho +{\rho }^{-1}$, its vertical semiaxis is $|\rho -{\rho }^{-1}|$, and its foci are $\pm 2$ because the squared semiaxes differ by $4$. Reciprocal radii give the same ellipse, traversed in opposite directions as $\theta$ increases.