= 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 $\zeta^2-a^2\ne0$ since $a<1$. Mapping these points into the physical plane gives
$$
\boxed{z_{\rm stag}=\pm\left[(1+a^2)\cos\alpha+i(1-a^2)\sin\alpha\right].}
$$
Equivalently, 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.
\Image[/past-exam-of-the-fluids-and-waves-course-of-the-university-of-oxford/2014-ellipse-flow.png]
{title=An oblique uniform flow around an ellipse has two mapped stagnation points}
{description=The example uses $a=0.65$, $U=1$ and $\alpha=30$ degrees. The marked points are $\pm((1+a^2)\cos\alpha,(1-a^2)\sin\alpha)$. Streamlines are level curves of the imaginary part of the mapped complex potential.}
Back to article page
