Set $S$ is the set of all complex numbers that can be squared to yield a real number. I solved by considering a complex number $z$ and setting the imaginary portion of its square equal to $0$.
$$\begin{align} & \Im\left(z^2\right)=0 \\ & \Im\left((a+bi)^2\right)=0,\,a,b\in\mathbb{R} \\ & 2ab=0 \\ & ab=0 \\ \end{align}$$
This means that $a=0$ or $b=0$:
$$\begin{align} & a+bi\to 0+bi=bi \\ & a+bi\to a+0i=a \\ \end{align}$$
This means that $S$ is a union of the set of real numbers and the set of pure imaginary numbers. Are there any alternative ways to go about proving this?
Another approach is to convert the complex number to its polar form.
Consider a complex number $z$:
$$\begin{align} & \Im\left(z^2\right)=0 \\ & \Im\left(m\cdot e^{2\theta i}\right)=0 \\ & 2m\cdot\cos(\theta)\sin(\theta)=0 \\ & \cos(\theta)\sin(\theta)=0 \\ \end{align}$$
This means that $(\cos(\theta)=0)$ + $(\sin(\theta)=0)$
$$\begin{align} & \cos(\theta)=0 \\ \implies& \theta=\frac{\pi}{2}+\pi n,\,n\in\mathbb{Z} \\ & \sin(\theta)=0 \\ \implies& \theta=\pi n,\,n\in\mathbb{Z} \\ \end{align}$$
This means that:
$$\theta=\frac{\pi}{2}n,\,n\in\mathbb{Z}$$
Therefore, $S$ is comprised only of numbers that form an angle measuring a multiple of $90^{\circ}$ on the complex plane. In other words, numbers on the real and imaginary axes of the complex plane.