= Solution
The <complex exponential> identity $\sin\theta=(e^{i\theta}-e^{-i\theta})/(2i)$ gives
$$
z=\frac{e^{2i\theta}-1}{2i},\qquad
\boxed{(\Re z)^2+(\Im z-\tfrac12)^2=\tfrac14.}
$$
Thus the <locus> is a <circle> of <radius> $1/2$ centered at $i/2$, traversed counterclockwise twice as $\theta$ runs through the specified interval. Its period as a parametrized <curve> is $\pi$. Near $\theta=0$, the <Taylor expansion> gives $\Re z=\theta+O(\theta^3)$ and $\Im z=\theta^2+O(\theta^4)$, so locally $\Im z=(\Re z)^2+O((\Re z)^4)$. There is \b[a smooth horizontal tangent, not a cusp]: $dz/d\theta=e^{2i\theta}$ is nonzero at each passage through the origin.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2021-locus-and-basis.png]
{title=A twice-traversed complex circle, a passive rotation of coordinate axes, and reflection in the line y=x/2}
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