 Solution

The Milne-Thomson circle theorem says that, when $f$ is a complex potential analytic inside and on $|z|=R$ and all its flow singularities are outside, the potential outside an impermeable circle isThe reflected expression is analytic in the exterior: conjugating a local power series turns its argument into $R^2/z$. It creates only interior image singularities. On $|z|=R$ we have $R^2/\bar{z}=z$, so $W=f+\bar{f}$ is real. The stream function is therefore constant on the circle, which is the impermeability condition. At infinity the reflected term is $\overline{f(0)}+O(z^{-1})$, so it changes the velocity by $O(z^{-2})$ and leaves the imposed far flow unchanged. These observations prove the theorem, up to the irrelevant additive potential constant and with no extra circulation introduced.

With $a=z_1$, $|a|>1$, and unit radius, it givesThe image dipole is at $1/\bar{a}$ with moment $-\bar{m}/{\bar{a}}^2$; writing the image term this way changes only an additive constant.