 Solution

The complex exponential identity $\sin \theta =(e^{i\theta }-e^{-i\theta })/(2i)$ givesThus 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 a smooth horizontal tangent, not a cusp: $dz/d\theta =e^{2i\theta }$ is nonzero at each passage through the origin.