Cube Roots Of Unity

In this chapter we will learn about an important application of complex numbers .

I. Introduction:

So far, we have learnt that $\sqrt[3]{1} = 1$ and $\sqrt[3]{-1} = -1$

But here, we will look at these problems from a different perspective. Let us take the case of $\sqrt[3]{1}$ and assume its value to be some unknown $x$.

Therefore,

$x = \sqrt[3]{1}$

$\Rightarrow x^3 = 1$

$\Rightarrow x^3 - 1 = 0$

The $LHS$ of the above relationship is a polynomial of order $3$, therefore, it should have $3$ roots.

The same holds true if we take $x = \sqrt[3]{-1}$

In the following two sections we will see that there are actually three roots for each of these two cases, one being a real number and the other two being complex numbers which we learnt .

II. Cube Roots Of Positive Unity:

We start with the relationship $x = \sqrt[3]{1}$

$\Rightarrow x^3 = 1$

$\Rightarrow x^3 - 1 = 0$

$\Rightarrow (x - 1)(x^2 + x + 1) = 0$

In the $LHS$ is a product of a linear term and a quadratic term.

Putting the linear term $x-1 = 0$ we get the real root, that is, $x = 1$

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If we take the quadratic term, and put it to $0$, we get:

$x^2 + x + 1 = 0$

$\Rightarrow x = \dfrac{-1 \pm \sqrt{1 - 4}}{2}$

$\Rightarrow x = \dfrac{-1 \pm \sqrt{-3}}{2}$

Substituting $\sqrt{-3} = \sqrt{3}i$, we get:

$x = \dfrac{-1 \pm \sqrt{3}i}{2}$

Therefore, the three cube roots of unity are $1, \dfrac{-1 + \sqrt{3}i}{2}$ and $\dfrac{-1 - \sqrt{3}i}{2}$

III. Cube Roots of Negative Unity:

Like before, we can set $x = \sqrt[3]{-1}$

$\Rightarrow x^3 = -1$

$\Rightarrow x^3 + 1 = 0$

$\Rightarrow (x + 1)(x^2 - x + 1) = 0$

Therefore, we get $x = -1$ as the real root, and solving the quadratic equation we get:

$x = \dfrac{1 \pm \sqrt{1 - 4}}{2}$

$\Rightarrow x = \dfrac{1 \pm \sqrt{3}i}{2}$

Therefore, the $3$ cube roots of $-1$ are, $-1, \dfrac{1 + \sqrt{3}i}{2}$ and $\dfrac{1 - \sqrt{3}i}{2}$

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IV. Properties Of Cube Roots Of Unity

The relationship between the two complex roots of unity can be derived by squaring any one of them.

Let us take the root $\dfrac{-1 + \sqrt{3}i}{2}$ of positive unity as $\omega$

Then,

$\omega^2 = \dfrac{(-1 + \sqrt{3}i)^2}{4}$

$\Rightarrow \omega^2 = \dfrac{1 - 2\sqrt{3}i + 3i^2}{4}$

$\Rightarrow \omega^2 = \dfrac{1 - 2\sqrt{3}i - 3 }{4}$

$\Rightarrow \omega^2 = \dfrac{-2 - 2\sqrt{3}i}{4}$

$\Rightarrow \omega^2 = \dfrac{-1 - \sqrt{3}i}{2}$

If we take $\omega$ as the other root $\dfrac{-1 - \sqrt{3}i}{2}$, then also we see that:

$\omega^2 = \dfrac{(-1 - \sqrt{3}i)^2}{4}$

$\Rightarrow \omega^2 = \dfrac{1 + 2\sqrt{3}i + 3i^2}{4}$

$\Rightarrow \omega^2 = \dfrac{1 + 2\sqrt{3}i - 3 }{4}$

$\Rightarrow \omega^2 = \dfrac{-2 + 2\sqrt{3}i}{4}$

$\Rightarrow \omega^2 = \dfrac{-1 + \sqrt{3}i}{2}$

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Now we can see that if $\omega$ is any one of the complex roots, then the other one is $\omega^2$. The same is true of the complex roots of negative unity.

V. Higher Roots Of Unity

We know from the Euler's Formula that:

$e^{ix} = \cos x + i \sin x$

Putting $x = \pi$, we get:

$e^{i\pi} = \cos \pi + i \sin \pi = -1 + 0 = -1$

$\therefore \left( e^{i\pi} \right)^{2k} = (-1)^{2k}$

$\Rightarrow e^{2ki \pi} = 1^k = 1$

$\therefore \sqrt[n]{1} = (1)^{\frac{1}{n}} = \left(e^{2ki\pi}\right)^{\frac{1}{n}}$

$\Rightarrow \sqrt[n]{1} = e^{\large{\frac{2ki\pi}{n}}}$