AMC 12 B, 2024

Let $\theta$ be the argument of $z$. Because $|z| = 2$ and the real part of $z$ is greater than $1$, it follows that $\theta$ is less than $60^\circ$. This ensures that the imaginary parts of $z^2$ and $z^3$ are positive and all the vertices of the quadrilateral other than $0$ lie in the upper half-plane. Thus the area of the quadrilateral is the sum of the areas of the triangle with vertices $0$, $z$, and $z^2$ and the triangle with vertices $0$, $z^2$, and $z^3$. See the figure.

Because the area of a triangle with side lengths $a$ and $b$ with included angle $\alpha$ is $\frac{1}{2}ab \sin \alpha$, the area of the quadrilateral must be
$$\frac{1}{2} (|z| \cdot |z|^2 \cdot \sin \theta + |z|^2 \cdot |z|^3 \cdot \sin \theta) = 20 \sin \theta = 15.$$
It follows that $\sin \theta = \frac{3}{4}$, and the imaginary part of $z$ is $2 \cdot \sin \theta = \frac{3}{2}$.