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For an analytic complex potential $f$ whose singularities lie outside $|z|=a$, the circle theorem adds the reflected potentialThe added term is analytic outside the circle wherever the reflected original is regular; in particular no new exterior singularities occur when $f$ is regular inside the circle. On $|z|=a$, the relation $a^2/\bar{z}=z$ makes $F=f+\bar{f}$ real. Thus its imaginary part, the stream function, is constant on the circle and the normal velocity vanishes. This proves the theorem, with any harmless additive constant left free. Its usual decay condition at infinity selects the surrounding flow inherited from $f$.

Write $a_1=z_1$ for the dipole position to distinguish it from the circle radius, now one. Substitution givesThe image singularity is at $1/{\bar{a}}_1$, strictly inside the solid circle. This exterior potential has the original dipole and an impermeable cylinder boundary.

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 dipole producesThe 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.

On the line $y=-x$ one has $\overline{iz}=z$. Hence the reflected term in the printed formula is $\overline{w(z)}$, and the total is real on the line. This proves that its stream function is constant there. The nested conjugation is important: it makes the added term analytic as a function of $z$.

For the planar potential dipole its image contribution isThusThe image is the reflected point $-i\bar{a}$, which lies in the opposite half-plane when the original dipole lies strictly inside the fluid half-plane. The wall itself is a limiting boundary, not an admissible position for the singularity.

A planar potential dipole at $a$ with complex moment $m$ has complex potential of a two-dimensional potential flow $w(z)=m/(z-a)$. Its complex velocity is $w^{\prime }(z)=-m/(z-a{)}^2$, interpreted as $v_x-i{v}_y$. Unlike the three-dimensional potential dipole, this is a two-dimensional singularity of the logarithmic-source potential. For example, two opposite source coefficients $s$ at $a\pm \delta /2$ have potential $s[\log (z-a-\delta /2)-\log (z-a+\delta /2)]$, whose leading term is $-s\delta /(z-a)$ as the separation tends to zero with $s\delta$ fixed. This gives the dipole as a limiting source–sink pair, with the complex sign and orientation included in $m$.

A transported planar potential dipole follows the surrounding regular velocity, so its position satisfies the displayed equation after its own singular term is removed. The conjugation comes from the convention $w^{\prime }=v_x-i{v}_y$. Reflected image parameters are held fixed when taking the spatial derivative, then evaluated at the instantaneous dipole position.

For a unit circular wall, the Milne-Thomson circle theorem gives image potential $\bar{m}z/(1-\bar{a}z)$ and therefore $\dot{\bar{a}}=\bar{m}/(1-|a{|}^2{)}^2$. For a horizontal straight wall the image is $\bar{m}/(z-\bar{a})$, giving $\dot{\bar{a}}=-\bar{m}/(a-\bar{a}{)}^2$. For the line $y=-x$, reflected conjugation gives image $-i\bar{m}/(z+i\bar{a})$ and hence $\dot{\bar{a}}=i\bar{m}/(a+i\bar{a}{)}^2$. In each case the image lies outside the fluid region and makes the wall a streamline; differentiation of the regular image proves the motion formula. Several compatible straight-wall reflections can be summed before the same operation is applied.