Find cube roots of unity and plot on Argand plane

Question

Find all the cube roots of unity and represent them on an Argand plane. Also verify that their sum is zero.

Solution — Step by Step

We need all complex numbers $z$ such that $z^3 = 1$ .

Rewriting: $z^3 - 1 = 0$ .

Factor: $(z - 1)(z^2 + z + 1) = 0$ .

From $z - 1 = 0$ : $z = 1$ .

So one cube root of unity is simply $\mathbf{1}$ (the real number 1).

From $z^2 + z + 1 = 0$ , use the quadratic formula:

z = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}

So the two complex cube roots of unity are:

\omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 - i\sqrt{3}}{2}

1 + \omega + \omega^2 = 1 + \frac{-1 + i\sqrt{3}}{2} + \frac{-1 - i\sqrt{3}}{2}

= 1 + \frac{(-1 + i\sqrt{3}) + (-1 - i\sqrt{3})}{2} = 1 + \frac{-2}{2} = 1 - 1 = 0 \checkmark

Convert to polar form to understand the geometry. All cube roots of unity have modulus 1 (they lie on the unit circle):

$1 = \cos 0° + i \sin 0°$ → at angle $0°$
$\omega = \cos 120° + i \sin 120°$ → at angle $120°$
$\omega^2 = \cos 240° + i \sin 240°$ → at angle $240°$

On the Argand plane, these three points form an equilateral triangle inscribed in the unit circle, with vertices equally spaced at $120°$ intervals.

Why This Works

The cube roots of unity are evenly distributed around the unit circle because the $n$ -th roots of unity always divide the unit circle into $n$ equal arcs. For $n = 3$ , we get three points separated by $360°/3 = 120°$ .

The key algebraic properties to memorise:

$\omega^3 = 1$
$1 + \omega + \omega^2 = 0$
$\omega^2 = \bar{\omega}$ (conjugates of each other)

Alternative Method — Using De Moivre’s Theorem

$1 = e^{i \cdot 0} = e^{i \cdot 2\pi} = e^{i \cdot 4\pi}$ (using periodicity of $e^{i\theta}$ )

z^3 = 1 \Rightarrow z = e^{i \cdot 2\pi k/3} \text{ for } k = 0, 1, 2

k=0: z = e^0 = 1

k=1: z = e^{i \cdot 2\pi/3} = \cos 120° + i\sin 120° = \frac{-1+i\sqrt{3}}{2} = \omega

k=2: z = e^{i \cdot 4\pi/3} = \cos 240° + i\sin 240° = \frac{-1-i\sqrt{3}}{2} = \omega^2

Common Mistake

Many students write down only two roots instead of three, thinking “cube root = one answer.” Remember: every non-zero complex number has exactly $n$ distinct $n$ -th roots. For cube roots, there are always 3 roots. The real root is 1; the complex roots are $\omega$ and $\omega^2$ .

The property $1 + \omega + \omega^2 = 0$ is used constantly in JEE problems. If you see an expression involving $1 + \omega + \omega^2$ , it equals zero. Similarly, $1 \cdot \omega \cdot \omega^2 = \omega^3 = 1$ (product of roots = 1). These two facts alone solve dozens of JEE Main and Advanced problems.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using De Moivre’s Theorem

Common Mistake

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