= Solution
{c}
For incompressible planar flow define the <streamfunction> by $u=\psi_y$, $v=-\psi_x$. A <velocity potential> satisfies $u=\phi_x$, $v=\phi_y$ when the flow is irrotational; the differential condition is $v_x-u_y=0$. The <complex potential of a two-dimensional potential flow> is $w=\phi+i\psi$, with derivative $w'=u-iv$.
For $w=m\log(z-z_0)/(2\pi)$, its derivative is $m/[2\pi(z-z_0)]$. This gives outward radial velocity $m/(2\pi r)$, whose flux across a circle is $2\pi r\cdot m/(2\pi r)=m$. It is therefore the stated line source.
Superposition gives
$$
w(z)=Uz+\frac m{2\pi}[\log z-\log(z-1)],\qquad
w'(z)=U-\frac m{2\pi z(z-1)}.
$$
For positive $U,m$, the stagnation points are
$$
\boxed{z_\pm=\frac{1\pm\sqrt{1+2m/(\pi U)}}2.}
$$
They lie on the real axis, one upstream of the source and the other downstream of the sink. Streamlines are level sets of $\psi=Uy+(m/2\pi)[\arg z-\arg(z-1)]$. The dividing <streamline> passes through both stagnation points and forms a <Rankine oval> enclosing the source and sink; outside, streamlines tend to uniform horizontal flow.
\Image[/past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2009-rankine-oval.png]
{title=A representative source-sink pair in a uniform stream, with two stagnation points and a dividing Rankine oval}
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