= Solution
For $q=W'=u-iv$, the <Blasius force theorem> states that the <force> per unit length exerted by a steady <incompressible flow> and <irrotational flow> on an impermeable body is
$$
X-iY=\frac{i\rho}{2}\oint_C q(z)^2\,dz,
$$
where $C$ runs counterclockwise around the body and excludes external flow singularities. The sign corresponds to the <force> on the body.
Let $c=a^2/b$ and $A=ma^2e^{-i\alpha}/b^2$. The <complex velocity> may be written
$$
q(z)=\frac{A}{(z-c)^2}+H(z),\qquad
H(z)=-\frac{me^{i\alpha}}{(z-b)^2}.
$$
Only the image point $c$ lies inside the circular contour. The term $A^2/(z-c)^4$ has zero <residue>; $H^2$ is regular there. In the cross term, the linear term of the <Taylor expansion> of $H$ supplies
$$
\operatorname{Res}_{z=c}q^2
=2AH'(c)
=\frac{4m^2a^2}{b^2(c-b)^3}
=-\frac{4m^2a^2b}{(b^2-a^2)^3}.
$$
The orientation factors have canceled. The <residue theorem> now gives
$$
X-iY=-\pi\rho\operatorname{Res}_{z=c}q^2,
\qquad
\boxed{X=\frac{4\pi\rho m^2a^2b}{(b^2-a^2)^3},\qquad Y=0}.
$$
Since $b>a$, $X>0$: the obstacle is pulled toward the doublet, independently of $\alpha$. This is a <force> per unit length perpendicular to the plane of flow.
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