= Solution
Take $\mathbf e=\mathbf e_x$ and the surface normal $\mathbf e_z$. Under reflection in the free surface, polar vectors transform by $M=\operatorname{diag}(1,1,-1)$ and an axial <torque> transforms by $\det(M)M$. A parallel <torque> therefore reverses sign, while the separation direction $\mathbf e_x$ does not. The <free-surface image of a rotlet dipole> is a dipole of strength $-D$ at $z=+h$.
The observation point at the cell is directly below the image, so $\mathbf e\cdot\mathbf r=0$ and the image <velocity> vanishes. A symmetry proof of the vanishing normal drift is also useful: reflection in the vertical plane containing $\mathbf e$ reverses the axial dipole moment, preserves the normal component of a polar <velocity>, and leaves the geometry unchanged. Linearity would make that same component change sign with $D$, forcing it to be zero. Thus \b[this singularity alone produces no attraction or repulsion].
Write $X,Y,Z$ relative to the image, so the cell is at $(0,0,-2h)$. The image field is
$$
\mathbf u_{\mathrm{im}}=\frac{3D}{8\pi\mu}
\frac{X(0,-Z,Y)}{(X^2+Y^2+Z^2)^{5/2}}.
$$
Although its value at the cell is zero, its <velocity> gradient is not. In particular, the <surface-induced yaw of a rotlet dipole> has
$$
\boxed{\omega_z=(\partial_Xu_Y-\partial_Yu_X)_{\mathrm{cell}}
=\frac{3D}{128\pi\mu h^4}\ne0\quad(D\ne0).}
$$
A spherical <torque>-free body rotates with the local fluid angular <velocity> $\Omega_z=\omega_z/2$. Its swimming direction therefore turns in the surface plane and, at fixed height and speed, it traces a circle with radius $V_{\mathrm{swim}}/|\Omega_z|$. The sign of $D$ sets the sense of turning in this convention. An elongated swimmer also responds to strain; its rotation coefficient depends on its shape and need not equal one half of the <vorticity>. When the <stresslet> is included, its changing height produces a changing turning rate rather than an exactly fixed-radius circle.
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