 Solution

Parameterize the unit circle by $\zeta =e^{i\theta }$. Under the Joukowski map,Consequently its image is the ellipsewith horizontal and vertical semi-axesBoth are positive because $0<a<1$. The horizontal axis is the major axis. The derivative $dz/d\zeta =1-a^2/{\zeta }^2$ is nonzero for $|\zeta |\ge 1$. Also, equality of two images implies $({\zeta }_1-{\zeta }_2)(1-a^2/({\zeta }_1{\zeta }_2))=0$; two distinct exterior points cannot satisfy the second factor since $|{\zeta }_1{\zeta }_2|>1>a^2$. Thus the exterior map is one-to-one, mapping the exterior of the unit circle conformally onto the exterior of the ellipse.