De Moivre's theorem — find cube roots of unity and plot on Argand plane

Question

Using De Moivre’s theorem, find all three cube roots of unity. Plot them on the Argand plane and verify that they form an equilateral triangle.

(JEE Main 2022, similar pattern)

Solution — Step by Step

We need to solve $z^3 = 1$ .

Write $1$ in polar form: $1 = \cos 0 + i\sin 0 = e^{i \cdot 0}$ .

But $1$ can also be written as $e^{i \cdot 2k\pi}$ for any integer $k$ . So:

z^3 = e^{i \cdot 2k\pi}

Taking the cube root:

z = e^{i \cdot 2k\pi/3} = \cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3}

For $k = 0, 1, 2$ (we need exactly 3 distinct roots):

$k = 0$ : $z_0 = \cos 0 + i\sin 0 = 1$

$k = 1$ : $z_1 = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$

$k = 2$ : $z_2 = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$

The cube roots of unity are conventionally written as $1, \omega, \omega^2$ where:

\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2} = e^{i \cdot 2\pi/3}

\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} = e^{i \cdot 4\pi/3}

Note that $\omega^2$ is the conjugate of $\omega$ , i.e., $\omega^2 = \bar{\omega}$ .

All three roots lie on the unit circle ( $|z| = 1$ ). They are separated by equal angles of $2\pi/3 = 120°$ .

The distance between any two adjacent roots:

|1 - \omega| = \left|1 + \frac{1}{2} - i\frac{\sqrt{3}}{2}\right| = \left|\frac{3}{2} - i\frac{\sqrt{3}}{2}\right| = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3}

Similarly, $|\omega - \omega^2| = \sqrt{3}$ and $|1 - \omega^2| = \sqrt{3}$ .

All three sides are equal ( $\sqrt{3}$ ), confirming an equilateral triangle inscribed in the unit circle.

Why This Works

De Moivre’s theorem states: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ . To find $n$ th roots, we reverse this: divide the angle by $n$ and add multiples of $2\pi/n$ to get all $n$ distinct roots.

The $n$ th roots of unity are always equally spaced on the unit circle, separated by angles of $2\pi/n$ . For $n = 3$ , this gives three points at $120°$ apart — hence the equilateral triangle.

Alternative Method — Key properties to remember

1 + \omega + \omega^2 = 0

\omega^3 = 1

\omega \cdot \omega^2 = \omega^3 = 1

These three properties are used constantly in JEE problems.

For JEE, the property $1 + \omega + \omega^2 = 0$ is tested in almost every paper — either directly or inside a larger problem. If you see a sum like $a + b\omega + c\omega^2$ , remember you can manipulate it using $\omega^2 = -1 - \omega$ . Also, $\omega^3 = 1$ means powers of $\omega$ cycle with period 3: $\omega^{100} = \omega^{99} \cdot \omega = (\omega^3)^{33} \cdot \omega = \omega$ .

Common Mistake

Students sometimes write $\omega = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ (positive real part). The correct value is $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ (negative real part). The angle is $120°$ , not $60°$ — the cosine of $120°$ is $-1/2$ , not $+1/2$ . Getting this sign wrong propagates through every calculation involving $\omega$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Key properties to remember

Common Mistake

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