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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 $b=(m+\sqrt{m^2+4U^2})/(2U)$.

Let $F_x,F_y$ denote the force per unit length exerted by the fluid on the cylinder, and let $\rho$ be the constant mass density. For a counterclockwise contour $C$ around the unit circle, the Blasius force theorem givesAn efficient use of the residue theorem finds the sum of the enclosed residues by subtracting the external pole at $b$ from the sum of all finite residues. At large $z$,The sum of all finite residues of $q^2$ is therefore $2U$. At $b$, write $q=(z-b{)}^{-1}+H(z)$, with $H$ a holomorphic function near $b$. ThenThusUsing $a=1/b$ and $1/(b-a)-1/b=1/[b(b^2-1)]$ givesThe vanishing vertical force also follows from reflection symmetry about the real axis. This force on a unit cylinder due to a uniform flow and an external source uses the counterclockwise orientation in the Blasius force theorem to fix the sign of the horizontal force.

For $q=W^{\prime }=u-iv$, the Blasius force theorem states that the force per unit length exerted by a steady incompressible flow and irrotational flow on an impermeable body iswhere $C$ runs counterclockwise around the body and excludes external flow singularities. The sign corresponds to the force on the body.

Let $c=a^2/b$ and $A=m{a}^2{e}^{-i\alpha }/b^2$. The complex velocity may be writtenOnly the image point $c$ lies inside the circular contour. The term $A^2/(z-c{)}^4$ has zero residue; $H^2$ is regular there. In the cross term, the linear term of the Taylor expansion of $H$ suppliesThe orientation factors have canceled. The residue theorem now givesSince $b>a$, $X>0$: the obstacle is pulled toward the doublet, independently of $\alpha$. This is a force per unit length perpendicular to the plane of flow.