 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 $|z|=a$ isReflection 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 $2\mathrm{Re}f(z)$: 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 $B/(a^2/z-b)=Bz/(a^2-bz)$, givingIts image pole is $c=a^2/b$, inside the cylinder. The complex velocity isNear $c$, the second term is $C(z-c{)}^{-2}$ with $C=B{a}^2/b^2$, 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 $2C{v}_0^{\prime }(c)$, soTake the contour counterclockwise around the cylinder, as in the supplied Blasius force theorem. The residue theorem gives $\frac{1}{2}i\rho \oint (w^{\prime }{)}^{2}dz=-\pi \rho \mathrm{Res}(w^{\prime }{)}^2$. ThereforeThe force points toward the dipole. Its independence of dipole orientation follows from the product $AB={\mu }^2$ in the sole residue contribution.