 Solution

The complex velocity vanishes where $e^{-i\alpha }{\zeta }^2-e^{i\alpha }=0$, provided its denominator is nonzero. Thus $\zeta =\pm e^{i\alpha }$. Both lie on the unit circle, and ${\zeta }^2-a^2\ne 0$ since $a<1$. Mapping these points into the physical plane givesEquivalently, the two stagnation points are the opposite points $(A\cos \alpha ,B\sin \alpha )$ and $(-A\cos \alpha ,-B\sin \alpha )$ on the ellipse. The conformal map preserves the zeros of the complex velocity here because its derivative is finite and nonzero on the boundary.