= Solution
On the surface, $u_r=0$ and $u_\theta=(3v/2)\sin\theta$. The only surface <stagnation points> are therefore
$$
\boxed{(0,0,R)\quad\text{and}\quad(0,0,-R).}
$$
There are no other <stagnation points> in the exterior: away from the axis the polar component is nonzero, and on the axis the radial component vanishes only at $r=R$.
Near the front point put $(X,Y,Z)=(x,y,z-R)$. Expanding the potential from <1/c/solution> gives
$$
\phi=-\frac{3vR}{2}+\frac{3v}{4R}(X^2+Y^2-2Z^2)+O(|\mathbf X|^3/R^2),
$$
so the linear <velocity> field is $(3vX/(2R),3vY/(2R),-3vZ/R)$. This agrees with <1/b/solution> for $W=3v/(4R)$.
At the rear point, using $Z=z+R$ instead, the quadratic term has the opposite sign: $W=-3v/(4R)$. Incoming axial flow and tangential outflow at the front become tangential inflow and axial outflow at the back. The local <streamline> geometry is unchanged while its direction reverses.
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