= Solution
The <Milne-Thomson circle theorem> states that if $f$ is an analytic complex potential with no <singularity> on or inside $|z|=a$, then
$$
w(z)=f(z)+\overline{f(a^2/\bar z)}
$$
makes $|z|=a$ a <streamline> while retaining the original exterior <singularities> and adding reflected <singularities> inside the <circle>. On the <circle> the two terms are conjugates, so the resulting <streamfunction> is zero there.
Put $A=me^{i\alpha}$ and $c=a^2/b$. Applying the formula to the exterior <planar potential doublet> gives
$$
\boxed{w(z)=\frac{me^{i\alpha}}{z-b}+\frac{me^{-i\alpha}z}{a^2-bz}.}
$$
Equivalently, up to an irrelevant constant,
$$
w(z)=\frac A{z-b}-\frac B{z-c},\qquad B=\frac{ma^2}{b^2}e^{-i\alpha}.
$$
The image <planar potential doublet> is inside the obstacle at $c<a$; the only physical <fluid> <singularity> is the prescribed one at $b$.
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