= Solution
Let the unit normal $\mathbf n$ point from the fluid toward the free surface, let $\mathbf t$ lie along the surface, and write the swimmer direction as
$$
\mathbf p=\cos\theta\,\mathbf t+\sin\theta\,\mathbf n.
$$
Reflecting the swimmer position across the plane and reflecting its orientation to
$$
\mathbf p^*=\cos\theta\,\mathbf t-\sin\theta\,\mathbf n
$$
constructs the <free-surface image of a force dipole>. At the surface the two dipoles have equal tangential velocity and opposite normal velocity. Their sum therefore satisfies the <no-penetration boundary condition>; tangential velocity is even across the plane and normal velocity is odd, so the tangential traction vanishes and the <stress-free boundary condition> is also satisfied.
The vector from the image to the swimmer is $\mathbf r=-2h\mathbf n$, for which $(\mathbf p^*\mathbin\cdot\mathbf r)^2=4h^2\sin^2\theta$. Evaluating the image <force-dipole flow> at the swimmer gives
$$
\boxed{
\mathbf U_{\rm surf}
=\frac{\mathcal P}{32\pi\mu h^2}
\left(1-3\sin^2\theta\right)\mathbf n
}.
$$
There is no tangential image velocity at the swimmer in this point-dipole approximation. With the convention that $\mathcal P>0$ is an extensile <pusher microswimmer>, a nearly parallel pusher is attracted toward the surface, whereas a nearly parallel contractile <puller microswimmer> is repelled. For $|\sin\theta|>1/\sqrt3$ the normal drift reverses because the image samples the axial rather than equatorial part of the dipolar flow.
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