 Solution

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.