= Solution
Across a straight impermeable wall the reflected potential is obtained by conjugating the potential after reflecting its argument. For the real axis the operation is $f(z)\mapsto\overline{f(\bar z)}$; the sum is real on that axis. For the imaginary axis it is $f(z)\mapsto\overline{f(-\bar z)}$. Applying both commuting reflections to the <planar potential dipole>[dipole] produces
$$
\boxed{F(z)=\frac m{z-a}+\frac{\bar m}{z-\bar a}-\frac{\bar m}{z+\bar a}-\frac m{z+a},\qquad a=z_1.}
$$
The other three poles are outside the first quadrant. The sum is invariant under each reflected-<conjugation> operation, hence real on each boundary axis. Both walls are therefore streamlines.
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