Question

Prove $1 + \omega + {\omega ^2} = 0$

Answer 1

Here, we will try to find the three cube roots of unity where one of them will be a real root and the other two will be imaginary roots. When we will add the squares of these three roots together, we will get a 0. Hence, by substituting the imaginary roots by omega and square of omega, we will be able to prove the given identity.

Formula Used:
$\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$
Determinant, $D = {b^2} - 4ac$
Quadratic formula, $z = \dfrac{{ - b \pm \sqrt D }}{{2a}}$

Complete step by step solution:
In order to prove the given identity,
Let us assume the cube root of unity or 1 as:
$\sqrt[3]{1} = z$
Cubing both sides, we get
$\Rightarrow 1 = {z^3}$
Subtracting 1 from both the sides, we get
$\Rightarrow {z^3} - 1 = 0$
Now, using the formula $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$
$\Rightarrow \left( {z - 1} \right)\left( {{z^2} + z + 1} \right) = 0$
Therefore, either $\left( {z - 1} \right) = 0$
$\Rightarrow z = 1$
Or, $\left( {{z^2} + z + 1} \right) = 0$
Comparing with $\left( {a{x^2} + bx + c} \right) = 0$
Here, $a = 1$, $b = 1$ and $c = 1$
Now, Determinant, $D = {b^2} - 4ac$
Hence, for $\left( {{z^2} + z + 1} \right) = 0$,
$D = {\left( 1 \right)^2} - 4 \times 1 = 1 - 4 = - 3$
Now, using quadratic formula,
$z = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Here, $a = 1$, $b = 1$and $c = 1$ and $D = - 3$
$\Rightarrow z = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2}$
This can be written as:
$\Rightarrow z = \dfrac{{ - 1 \pm \sqrt 3 i}}{2}$
Therefore, the three cube roots of unity are:
$1$, $\dfrac{{ - 1}}{2} + \dfrac{{\sqrt 3 i}}{2}$ and $\dfrac{{ - 1}}{2} - \dfrac{{\sqrt 3 i}}{2}$
Now, according to the property, the sum of these three cube roots of unity will be equal to 0.
Here, $\omega$ represents the imaginary cube roots.
$\Rightarrow 1 + \dfrac{{ - 1}}{2} + \dfrac{{\sqrt 3 i}}{2} + \dfrac{{ - 1}}{2} - \dfrac{{\sqrt 3 i}}{2} = 1 - 1 + 0 = 0$
$\Rightarrow 1 + \omega + {\omega ^2} = 0$
Here, $\omega = \dfrac{{ - 1}}{2} + \dfrac{{\sqrt 3 i}}{2}$
And, ${\omega ^2} = \dfrac{{ - 1}}{2} - \dfrac{{\sqrt 3 i}}{2}$
Therefore,
$1 + \omega + {\omega ^2} = 0$
Hence, proved

Note:
We should know that the cube root of any number is that number which when multiplied two times by itself gives the same number. Hence, the cube root of unity means that when a number is raised to the power 3, it gives the answer 1. The values of cube roots of unity are. Here, one root is a real number and the other two are the imaginary or complex numbers. Hence, we represent the imaginary numbers as $\omega$ and ${\omega ^2}$