Roots of the equation $x^4-2x^2+4=0$ form which shape?

Question

Solving the quadratic equation

$$x^2=1\pm i\sqrt 3$$

$$x^2=-2\omega, -2\omega^2$$ where $\omega$ is a cube root of unity

Then $$x=\pm i\sqrt 2 \omega^2, \pm i \sqrt 2 \omega$$

Now $|x|=\sqrt 2$, so the points should lie on a circle of radius $\sqrt 2$, but how can we tell if it’s a square, rectangle or rhombus ?

Answer 1

Let $x_1 = +i\sqrt 2 \omega^2, x_2 = + i \sqrt 2 \omega, x_3 = -i\sqrt 2 \omega^2, x_4 = -i\sqrt 2 \omega$.

As, $x_1 = -x_3$ and $x_2 = -x_4$, so they must lie opposite to each other.

Now, $x_1 - x_2 = x_4 - x_3 = i\sqrt2 (\omega^2 - \omega) = \sqrt6$, thus parallel to real axis.

Also, $x_1 - x_3 = x_2 - x_4 = i\sqrt2 (\omega^2 + \omega) = -i\sqrt2$, thus parallel to imaginary axis.

Therefore the figure is a rectangle as all angles are $90\deg$ and length of height and width are different.

Answer 2

$\omega$ and $\omega^2$ have $120^{\circ}$ between them, and they have the same real part. Multiplying each by $i$, the two results still have $120^{\circ}$ between them, and now the same imaginary part. These are in quadrants III and IV. The other two roots are their negatives/conjugates.

The roots form a rectangle, oriented with the axes, with the height less than the width.

Answer 3

Hint: write in polar coordinates and consider that $i=e^{i\pi/2}$, which is a rotation by 90 degrees.