Introduction to Cube Roots of Unity
In the realm of complex numbers, the 'cube roots of unity' refer to the three complex numbers that satisfy the equation $x^3 = 1$. While we often think of $1$ as the only cube root of $1$, this is only true in the real number system. In the complex plane, there are two additional imaginary roots.
Solving for the Cube Roots
To find the cube roots of unity, we solve the equation: $$x^3 = 1$$
Subtract 1 from both sides: $$x^3 - 1 = 0$$
Using the difference of cubes formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, we get: $$(x - 1)(x^2 + x + 1) = 0$$
This gives us two cases:
- $x - 1 = 0 \implies x = 1$
- $x^2 + x + 1 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for the second part: $$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$$
Thus, the three cube roots are:
- $1$
- $\omega = \frac{-1 + i\sqrt{3}}{2}$
- $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$
Proving 1 + ω + ω² = 0
We are given that $\omega$ is an imaginary cube root of unity. This means $\omega$ must satisfy the quadratic equation $x^2 + x + 1 = 0$ that we derived above.
Since $\omega$ is a root, substituting it into the equation gives: $$\omega^2 + \omega + 1 = 0$$
Rearranging the terms, we arrive at the identity: $$1 + \omega + \omega^2 = 0$$
Alternative Algebraic Proof
Alternatively, consider the sum of the roots of the equation $x^3 - 1 = 0$, which is $x^3 + 0x^2 + 0x - 1 = 0$. According to Vieta's formulas, for a polynomial $ax^n + bx^{n-1} + \dots = 0$, the sum of the roots is given by $-\frac{b}{a}$. Here, $a=1$ and $b=0$, so the sum of the roots ($1 + \omega + \omega^2$) is: $$-\frac{0}{1} = 0$$
Hence, it is proved that 1 + ω + ω² = 0.