Squares in the Complex plane In the complex plane, $z,$ $z^2,$ $z^3$ form, in some order, three of the vertices of a non-degenerate square. What are all the possible values of the area of the square?
I tried doing $z=a+bi$ so $z^2=a^2-b^2+2ab i$ $\sqrt{(a^2-b^2-a)^2+(2ab-b)^2}=\sqrt{(a^3-3ab^2-a)^2+(3a^2-b^3-b)^2}$ but got nowhere. Help!
 A: You have three choices, depending which pair of numbers is on the diagonal. The only two things I will use is that the lengths of sides is the same, and that the length of the diagonal is $\sqrt2$ times the length of the sides. Note that $z$ can't be any real number, since then $z$, $z^2$, and $z^3$ would be on the same line.

*

*Case $z$ and $z^3$ are on the diagonal.
It means $$|z^2-z|=|z^3-z^2|$$I rewrite this as $$|z||z-1|=|z|^2|z-1|$$Since we know that $z$ is not real, $|z|\ne 0$ and $|z-1|\ne 0$. Then $|z|=1$ and therefore $z, z^2, z^3$ will be on the unit circle, so the area of the square is $A=2$.

*Case $z$ and $z^2$ are on the diagonal.
Similar to before $$|z^3-z|=|z^3-z^2|\\|z||z-1||z+1|=|z|^2|z-1|$$
Therefore $$|z+1|=|z|$$
This is not enough, so I will use the length of the diagonal as well:
$$|z^2-z|=\sqrt 2 |z^3-z^2|$$ This yields $$|z|=\frac 1{\sqrt 2}$$
So $z$ is at the intersection of two circles of radius $1/\sqrt 2$, one centered on the origin, the other one on $-1+0i$. It's easy to see that $$z=-\frac 12\pm i\frac 12$$
For $z$ in the upper half, you get $$z^2=-\frac i2\\z^3=\frac 14+i\frac14$$
I've let you verify that the sides are equal, the diagonal is $\sqrt2$ times the sides and that the area is $A=\frac 58$.

*Case $z^3$ and $z^2$ are on the diagonal.
$$|z^3-z|=|z^2-z|\\|z+1|=1$$
For the diagonal $$|z^3-z^2|=\sqrt|z^2-z|\\|z|=\sqrt 2$$
So $z$ is at the intersection of a circle of radius $1$ centered on $-1+0i$, and a circle of radius $\sqrt 2$ centered on the origin. So $$z=-1\pm i$$
Using the $z$ in the upper half of the plane you get $$z^2=-2i$$
Therefore $z^2-z=1-3i$, so the area is $A=10$
