= Solution
For a planar <incompressible flow>, a <stream function> satisfies $u=\psi_y$, $v=-\psi_x$. A <velocity potential> satisfies $u=\phi_x$, $v=\phi_y$ where the flow is irrotational; such a potential exists locally, and globally on a simply connected nonsingular region. The <irrotational flow> condition is $v_x-u_y=0$. Where both exist, the <Cauchy-Riemann equations> make the <complex potential of a two-dimensional potential flow>[complex potential] $w=\phi+i\psi$ analytic with $w'=u-iv$.
For the logarithmic potential, write $z-z_0=re^{i\theta}$. Then
$$
\phi=\frac m{2\pi}\log r,\qquad \psi=\frac m{2\pi}\theta,
\qquad u_r=\frac m{2\pi r},\quad u_\theta=0.
$$
The outward <volume flux> per unit span across a circle is $2\pi r u_r=m$. Hence it is a <line source> of strength $m$. The potential may use local logarithm branches; its velocity is single-valued away from the source.
By <superposition>, the full flow has
$$
w(z)=Uz+\frac m{2\pi}\log z-\frac m{2\pi}\log(z-1),\qquad
w'(z)=U-\frac m{2\pi z(z-1)}.
$$
For positive $m$ and $U$, the <stagnation points> are
$$
\boxed{z_\pm=\frac{1\pm\sqrt{1+2m/(\pi U)}}2}.
$$
They are real, one upstream of the source and one downstream of the sink. Away from the singularities, <streamlines> are level curves of
$$
\psi(x,y)=Uy+\frac m{2\pi}\bigl(\arg z-\arg(z-1)\bigr).
$$
The zero level through the two stagnation points contains an upper and a lower separating curve, symmetric about the real axis and about $x=1/2$. Together they bound a <Rankine oval>. Flow inside goes from source to sink; the external stream divides around this oval and tends to horizontal flow far away. This also specifies the sketch for every positive parameter ratio; the figure illustrates one ratio.
\Image[/past-exam-of-the-options-i-course-of-the-university-of-oxford/2009-rankine-oval.png]
{title=Source–sink flow in a uniform stream}
{description=Arrows follow the velocity; the bold separating streamline passes through both stagnation points. The illustrated ratio is $m/(2\pi U)=1/4$.}
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