Past exam of the options II course of the University of Oxford / 2012 / K / K3 / c

The complex velocity obtained from part (b) isFor a nonzero planar potential doublet, $B/A=(a^2/b^2)e^{-2i\alpha }$. Its zeros therefore satisfyand henceThe denominators are nonzero because $a/b<1$. For either sign,so $|z_{\pm }|=a$. They are distinct because $t_{+}\ne t_{-}$ and $t\mapsto (c-bt)/(1-t)$ is injective, its determinant being $c-b\ne 0$. Clearing the velocity denominators gives a quadratic with these two roots, so there are no further finite stagnation points. Thus both stagnation points lie on the obstacle boundary.