= Solution
{c}
Writing $z=re^{i\theta}$ gives $\phi=\Gamma\theta/(2\pi)$ and $\psi=-\Gamma\log r/(2\pi)$. Constant <streamfunction> means constant radius, so streamlines are circles. The tangential velocity is $v_\theta=\Gamma/(2\pi r)$, and its anticlockwise <circulation> is $\int_0^{2\pi}v_\theta r\,d\theta=\Gamma$.
Use an opposite vortex at $z=-1$ as the image of the vortex at one:
$$
\boxed{w(z)=-\frac{i\Gamma}{2\pi}[\log(z-1)-\log(z+1)].}
$$
On $z=iy$ the two distances are equal, so $\psi=0$ and the wall is a <streamline>. Its velocity follows from $w'=-i\Gamma/[\pi(z^2-1)]$: at the wall $u=0$ and $v=-\Gamma/[\pi(1+y^2)]$.
The <Bernoulli equation> gives $p-p_\infty=-\rho\Gamma^2/[2\pi^2(1+y^2)^2]$. Relative to the ambient pressure force, the fluid's excess force on the wall is $-\int(p-p_\infty)dy$ in the positive $x$ direction. Using the supplied integral,
$$
\boxed{\mathbf F_{\rm vortex}=\frac{\rho\Gamma^2}{4\pi}\,\mathbf e_x\quad\text{per unit width}.}
$$
Thus the wall is attracted toward the fluid and vortex. Subtracting the uniform ambient pressure isolates the finite force due to the vortex.
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