 Solution

Keep the prescribed orientation $\alpha$ and strength $m$ fixed, and now let the doublet position $b(t)$ be complex. The Milne-Thomson circle theorem applies instantaneously withWhen taking its spatial derivative, $b(t)$ is a parameter. Removing the doublet's own singular term leaves the regular complex velocity at its position:Because $q=u-iv$, the stipulated image-induced motion of a planar doublet obeys $\dot{b}=\overline{q_{\mathrm{reg}}(b)}$, not $q_{\mathrm{reg}}(b)$ itself. HenceIts direction is constant. Indeed $\mathrm{Im}(b{e}^{-i\alpha })$ is constant and, since $b(0)=b_0>a$ is real, equals $-b_0\sin \alpha$. Writing $b=r{e}^{i\theta }$ in polar coordinates gives the path equationwhere the quotient is used when $\sin \alpha \ne 0$ and the branch through $(r,\theta )=(b_0,0)$ with $r>a$ is selected. It is the straight line through $b_0$ parallel to $e^{i\alpha }$. An unambiguous parametrization, also covering the degenerate polar cases, isIf $\sin \alpha =0$, the path is radial along the positive real axis, with $\theta =0$ and changing $r$; it cannot be represented as a single-valued $r(\theta )$. The sign of $m$ selects which direction along the line is followed. These paths apply only while the singularity remains outside the obstacle; the model prescribes no orientation evolution.