If $z_1$, $z_2 \in \mathbb{C}$ then $|z_1^2 - z_2^2| = |z_1-z_2|\cdot |z_1 + z_2|$. If moreover $|z_1| = |z_2|$ then $|z_1^2 - z_2|^2$ equals twice the area of the rhombus on the vectors $z_1$, $z_2$.

So now consider a set of points $M = \{z_n \ | z_n \in \mathbb{Q}(i), |z_n| = 1\}$. The set $\{z_n^2 \ | \ z_n \in M \}$ will work ( can you see why?)

Note that $M = \{-1\} \cup \{ \frac{1-t^2}{1+t^2} + i \frac{2 t}{1+t^2} \ |\ t \in \mathbb{Q} \}$

If $z_1$, $z_2 \in \mathbb{C}$ then $|z_1^2 - z_2^2| = |z_1-z_2|\cdot |z_1 + z_2|$. If moreover $|z_1| = |z_2|$ then $|z_1^2 - z_2|^2$ equals twice the area of the rhombus on the vectors $z_1$, $z_2$.

So now consider a set of points $M = \{z_n \ | z_n \in \mathbb{Q}(i), |z_n| = 1\}$. The set $\{z_n^2 \ | \ z_n \in M \}$ will work ( can you see why  ? $\ $ $\bf{A:}$ The area of the said rhombus is rational)

Note that $M = \{-1\} \cup \{ \frac{1-t^2}{1+t^2} + i \frac{2 t}{1+t^2} \ |\ t \in \mathbb{Q} \}$