Introduction to Cube Roots of Unity
In the field of complex numbers, the cube roots of unity are the solutions to the polynomial equation:
$$z^3 = 1$$
While we are familiar with the real number $1$ as a solution, there are two additional complex numbers that also satisfy this equation.
Deriving the Roots
To find the roots, we rearrange the equation:
$$z^3 - 1 = 0$$
Using the algebraic identity for the difference of two cubes, $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, we get:
$$(z - 1)(z^2 + z + 1) = 0$$
This gives us two cases:
- Case 1: $z - 1 = 0 \implies z = 1$
- Case 2: $z^2 + z + 1 = 0$
Applying the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to the second equation:
$$z = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$$
Thus, the three cube roots of unity are:
- $1$
- $\omega = \frac{-1 + i\sqrt{3}}{2}$
- $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$
Key Properties
The cube roots of unity possess fascinating properties that simplify complex algebraic problems:
The Sum of Roots is Zero: $$1 + \omega + \omega^2 = 0$$ This follows directly from the quadratic equation $z^2 + z + 1 = 0$.
The Product of Roots: $$\omega^3 = 1$$ If you multiply $\omega$ by $\omega^2$, you obtain $1$. More generally, $\omega^{3n} = 1$ for any integer $n$.
Geometric Interpretation: When plotted on the complex plane (Argand diagram), these three roots lie on a unit circle. They form the vertices of an equilateral triangle centered at the origin, with each root separated by an angle of $120^\circ$ ($2\pi/3$ radians).
Complex Conjugates: The two non-real roots are complex conjugates of each other. That is, $\bar{\omega} = \omega^2$ and $\overline{\omega^2} = \omega$.