= Solution
<Milne-Thomson's circle theorem> inserts an impermeable circular boundary into a two-dimensional <irrotational> <incompressible flow>. If $f(z)$ is analytic in a neighborhood of the <closed disc> and its flow singularities lie outside, the new complex potential outside $|z|=a$ is
$$
\boxed{w(z)=f(z)+\overline{f(a^2/\bar z)}.}
$$
Reflection and coefficient conjugation make the second term analytic outside the <circle>, with its image singularities inside. On $|z|=a$, $a^2/\bar z=z$, so the two terms sum to $2\operatorname{Re}f(z)$: the <streamfunction> is constant there. Thus no <velocity> crosses the <circle>. At infinity the image term tends to the constant $\overline{f(0)}$, which has no effect on <velocity>, and the far-field flow is retained. This proves the construction, with an additive potential constant immaterial.
Put $A=\mu e^{i\alpha}$, $B=\mu e^{-i\alpha}$. For the dipole the reflected potential is $B/(a^2/z-b)=Bz/(a^2-bz)$, giving
$$
\boxed{w(z)=\frac{A}{z-b}+\frac{Bz}{a^2-bz}.}
$$
Its image pole is $c=a^2/b$, inside the <cylinder>. The <complex velocity> is
$$
w'(z)=-\frac A{(z-b)^2}+\frac{Ba^2}{(a^2-bz)^2}.
$$
Near $c$, the second term is $C(z-c)^{-2}$ with $C=Ba^2/b^2$, while $v_0(z)=-A/(z-b)^2$ is analytic there. The square of the pole term has no <residue>, nor does $v_0^2$. The cross term has <residue> $2Cv_0'(c)$, so
$$
\operatorname{Res}_{z=c}(w')^2
=\frac{4ABa^2b}{(a^2-b^2)^3}
=-\frac{4\mu^2a^2b}{(b^2-a^2)^3}.
$$
Take the contour counterclockwise around the <cylinder>, as in the supplied <Blasius force theorem>. The <residue theorem> gives $\tfrac12i\rho\oint(w')^2dz=-\pi\rho\operatorname{Res}(w')^2$. Therefore
$$
\boxed{F_x=\frac{4\pi\rho\mu^2a^2b}{(b^2-a^2)^3},\qquad F_y=0.}
$$
The force points toward the dipole. Its independence of dipole orientation follows from the product $AB=\mu^2$ in the sole <residue> contribution.
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