= Solution
The <chain rule> gives
$$
\boxed{U-iV=\frac{dG}{dz}=\frac{d\psi/d\zeta}{df/d\zeta}=\frac{\cos z}{2+\sin z}.}
$$
To make the directions explicit, set $z=x+iy$ and $D=(2+\sin x\cosh y)^2+\cos^2x\sinh^2y$. Multiplying by the <complex conjugate> of the denominator gives the <logarithmic flow in a sine-mapped half-strip>:
$$
\boxed{U=\frac{\cos x(2\cosh y+\sin x)}D,\qquad V=\frac{\sinh y(\cosh y+2\sin x)}D.}
$$
In the open strip $U>0$. The vertical component is negative near the lower left wall where $\cosh y+2\sin x<0$, and positive above that curve. On the bottom the flow is rightward. On the right wall it is upward, whereas on the left wall it points downward below $z_*$ and upward above $z_*$. The field is tangent to the boundaries except at the source and vanishes at the two bottom corners. At large $y$, $U-iV\to-i$, so it tends to a uniform upward field $(0,1)$.
Near $z_*$, $G'(z)\sim1/(z-z_*)$, showing a radial boundary source. The <streamlines> are the curves $\operatorname{Im}G=\arg(2+\sin z)=\text{constant}$; the figure shows their orientation without reusing the exam's graphics.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-q7-sine-map-flow.png]
{title=Radial flow in the upper half-plane and its image in a vertical half-strip}
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