= Solution
The <Milne-Thomson circle theorem> says that, when $f$ is a <complex potential of a two-dimensional potential flow>[complex potential] analytic inside and on $|z|=R$ and all its flow singularities are outside, the potential outside an impermeable circle is
$$
W(z)=f(z)+\overline{f(R^2/\bar z)}.
$$
The 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 gives
$$
\boxed{W(z)=\frac{m}{z-a}+\frac{\bar m z}{1-\bar a z}}.
$$
The 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.
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