= Solution
The <complex potential of a two-dimensional potential flow> gives
$$
q(z):=W'(z)
=-\frac{me^{i\alpha}}{(z-b)^2}
+\frac{ma^2e^{-i\alpha}}{(bz-a^2)^2}.
$$
At a <stagnation point>, $q=0$. For $m\ne0$, taking the two square-root alternatives gives
$$
bz-a^2=\sigma a e^{-i\alpha}(z-b),\qquad \sigma\in\{+1,-1\}.
$$
Therefore
$$
\boxed{z_\sigma
=a\,\frac{a-\sigma b e^{-i\alpha}}{b-\sigma a e^{-i\alpha}},
\qquad \sigma=\pm1}.
$$
The denominators are nonzero because $b>a$. Moreover
$$
|a-\sigma b e^{-i\alpha}|^2
=a^2+b^2-2\sigma ab\cos\alpha
=|b-\sigma a e^{-i\alpha}|^2,
$$
so both <stagnation points> satisfy $|z_\sigma|=a$. They are distinct: equality would require $2e^{-i\alpha}(a^2-b^2)=0$. Neither is a singularity of $q$. Conversely, the numerator of $q$ is a quadratic polynomial with nonzero leading coefficient, so these are all its finite zeros. Thus there are exactly two boundary <stagnation points> and none in the open fluid region. If $m=0$, the flow instead vanishes everywhere, which is the degenerate exception to the assertion.
Back to article page
