= Solution
In the <analytic continuation> inside the <circle>, only the image location $c=a^2/b$ is singular; $b$ lies outside. Squaring the velocity gives
$$
(w')^2=\frac{A^2}{(z-b)^4}-\frac{2AB}{(z-b)^2(z-c)^2}+\frac{B^2}{(z-c)^4}.
$$
The last term has no <residue>, and the middle term has <residue>
$$
\operatorname{res}((w')^2;c)=\left.\frac{d}{dz}\left(-\frac{2AB}{(z-b)^2}\right)\right|_{z=c}
=\frac{4AB}{(c-b)^3}
=-\frac{4m^2a^2b}{(b^2-a^2)^3},
$$
because $AB=m^2a^2/b^2$ and $c-b=-(b^2-a^2)/b$. The <residue theorem> and part (a) therefore yield
$$
\boxed{F_x-iF_y=-\pi\rho\operatorname{res}((w')^2;c)=\frac{4\pi\rho m^2a^2b}{(b^2-a^2)^3}.}
$$
Thus $F_y=0$ and $F_x>0$. \b[The <force> points towards the <planar potential doublet> and is <independent> of its orientation], since the factors $e^{i\alpha}$ and $e^{-i\alpha}$ cancel in $AB$.
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