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In a planar potential flow, superpose a positive-$x$ uniform flow of speed $U$ and a line source of strength $Q=2\pi m$ at the real point $b>1$. Introduce an impermeable unit circular cylinder with zero circulation. The Milne-Thomson circle theorem gives the complex potential of a two-dimensional potential flowThe displayed horizontal force per unit length acts on the cylinder, and $F_y=0$ by reflection symmetry. To prove it, let $q=w^{\prime }$ be the complex velocity. At infinity, $q=U+m/z+O(z^{-2})$, so the sum of all finite residues of $q^2$ is $2Um$. At the external pole $b$, write $q=m/(z-b)+H(z)$; thenThe residue theorem and the counterclockwise Blasius force theorem give $F_x-i{F}_y=2\pi \rho m[H(b)-U]$, proving the formula. With $m>0$ and $U>0$, its sign changes at $b=(m+\sqrt{m^2+4U^2})/(2U)$.

A line sink is a planar line source with negative volume flux: fluid moves radially inward towards its location. For withdrawal rate $S>0$, its complex potential of a two-dimensional potential flow is $-S\log (z-b)/(2\pi )$.

In a planar incompressible flow, a line source injects volume flux $Q$ per unit length perpendicular to the plane. Its radial velocity is $u_r=Q/(2\pi r)$, since integrating the outward velocity around a circle gives $2\pi r{u}_r=Q$. Its complex potential of a two-dimensional potential flow is $w=Q\log (z-b)/(2\pi )$ and its complex velocity is $w^{\prime }=Q/[2\pi (z-b)]$. Here $Q>0$ means outward flow; a line sink has negative $Q$. Away from $b$, this is an irrotational flow as well as an incompressible flow.

Since $b>1$, the denominator inis positive. The positive root of $U{b}^2-b-U=0$ isThe other root is negative, so the horizontal force points in the positive $x$ direction for $1<b<b_{\ast }$, vanishes at $b=b_{\ast }$, and points in the negative $x$ direction for $b>b_{\ast }$. In particular, the nearby line source pulls the cylinder towards itself as $b\downarrow 1$, while at large $b$ the horizontal force is $F_x\sim -2\pi \rho U/b^2$.

Use $w=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so its complex velocity is $q=w^{\prime }=u-iv$. The incident uniform flow and line source haveThe coefficient of the complex logarithm is one because a line source of strength $Q$ has complex potential of a two-dimensional potential flow $(Q/(2\pi ))\log (z-b)$, and here $Q=2\pi$. Applying the Milne-Thomson circle theorem to the unit circle givesSince $1/z-b=-b(z-1/b)/z$, locally consistent branches of the complex logarithm giveChanges of branch only alter the constant in a local expression; $w^{\prime }$ is single valued away from its singularities. The added terms represent a two-dimensional potential doublet at the origin, an image line source at $1/b$, and a compensating line sink at the origin. The image line source and line sink have equal and opposite strengths, so no net volume flux enters the cylinder.

On $|z|=1$, the Milne-Thomson circle theorem makes $w=f(z)+\overline{f(z)}$ up to a constant. Consequently the streamfunction is constant there, establishing the no-penetration boundary condition. All line source strengths are real, and the two-dimensional potential doublet has zero circulation, so the circulation around the cylinder is zero. There is no additional line vortex term.