 Solution

The circle theorem says that if a complex potential $f$ has no singularities on or inside $|z|=a$ (and has the required analytic continuation there), thenadds image singularities inside the circle and makes it a streamline. On the circle $a^2/\bar{z}=z$, so $w=f(z)+\overline{f(z)}$ is real and its imaginary part is constant. This gives zero normal velocity. The reflected term is holomorphic where defined, as follows from its power series in $1/z$, and its singularities are precisely the reflected ones; the external original singularities are retained. This proves the construction, with an additive constant and any independently prescribed circulation handled separately.

For the fluid vortex use $f(z)=-i\Gamma \log (z-b)/(2\pi )$. Reflecting gives an opposite image at $d=a^2/b$ and also a term $-i\Gamma \log z/(2\pi )$, up to constants. The zero-circulation circle-theorem construction has both these terms. To impose the specified circulation $-\Gamma$ about the obstacle, add $+i\Gamma \log z/(2\pi )$, canceling the central term. ThusIt has the stated circulation and the same impermeable boundary, since the added central-fluid vortex streamfunction is constant on the circle.

For the given Blasius integral take the obstacle contour counterclockwise. Inside it the sole singularity ofis at $d$. Its residue is ${\Gamma }^2/[2{\pi }^2(d-b)]$, from the cross term. The residue theorem then givesTherefore $\boxed{F_x=\rho {\Gamma }^{2}b/[2\pi (b^2-a^2)],F_y=0}$. The force on the obstacle points toward the external fluid vortex.