Cube Roots of Unity ($\omega$) — Power Plays for JEE Main

Introduction

The cube roots of unity are among the most frequently tested concepts in JEE Main Complex Numbers. Mastering the properties of $\omega$ can help you solve problems in seconds that would otherwise take minutes.

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Definition and Values

The cube roots of unity are the solutions to $z^3 = 1$.

Factoring: $z^3 - 1 = (z - 1)(z^2 + z + 1) = 0$

The Three Roots:

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Essential Properties to Memorize

Property 1: Sum Equals Zero

$1 + \omega + \omega^2 = 0$

Corollary: $\omega + \omega^2 = -1$

Property 2: Product Equals One

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

Property 3: Conjugate Relationship

$\bar{\omega} = \omega^2 \quad \text{and} \quad \bar{\omega^2} = \omega$

Property 4: Cyclicity

$\omega^3 = 1, \quad \omega^4 = \omega, \quad \omega^5 = \omega^2, \quad \omega^6 = 1, \ldots$

General Rule: $\omega^n = \omega^{n \mod 3}$

Property 5: Roots of $x^2 + x + 1 = 0$

The equation $x^2 + x + 1 = 0$ has roots $\omega$ and $\omega^2$.

Property 6: Geometric Representation

$1$, $\omega$, $\omega^2$ form an equilateral triangle on the unit circle in the Argand plane.

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The "$\sqrt{3}$ and $1$" Trick

Whenever you see $\sqrt{3}$ and $1$ combined in a complex number problem, think $\omega$ or $\omega^2$.

Conversions: $\omega = \frac{-1 + i\sqrt{3}}{2} \implies -1 + i\sqrt{3} = 2\omega$ $\omega^2 = \frac{-1 - i\sqrt{3}}{2} \implies -1 - i\sqrt{3} = 2\omega^2$

Also useful: $1 + i\sqrt{3} = -2\omega^2$ $1 - i\sqrt{3} = -2\omega$

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JEE Main Previous Year Questions

PYQ 1 (JEE Main 2017)

Let $\omega$ be a complex number such that $2\omega + 1 = z$ where $z = \sqrt{-3}$. If $\begin{vmatrix} 1 & 1 & 1 \\ 1 & -\omega^2-1 & \omega^2 \\ 1 & \omega^2 & \omega^7 \end{vmatrix} = 3k$ then $k$ is equal to: (1) $1$ (2) $-z$ (3) $z$ (4) $-1$

Solution:

Given: $2\omega + 1 = i\sqrt{3}$

So $\omega = \frac{-1 + i\sqrt{3}}{2}$, which is the cube root of unity!

Using properties:

• $\omega^3 = 1 \implies \omega^7 = \omega^6 \cdot \omega = \omega$
• $-\omega^2 - 1 = \omega$ (since $1 + \omega + \omega^2 = 0$)

The determinant becomes: $\begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix}$

Apply $R_1 \to R_1 + R_2 + R_3$: $\begin{vmatrix} 3 & 1+\omega+\omega^2 & 1+\omega+\omega^2 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix} = \begin{vmatrix} 3 & 0 & 0 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix}$

Expanding along $R_1$: $= 3(\omega \cdot \omega - \omega^2 \cdot \omega^2) = 3(\omega^2 - \omega^4) = 3(\omega^2 - \omega) = 3\omega^2(1 - \omega^{-1})$

Actually, let's compute directly: $= 3(\omega^2 - \omega^4) = 3(\omega^2 - \omega)$

Since $\omega^2 - \omega = \omega^2 - \omega$, and we know: $\omega - \omega^2 = \frac{(-1+i\sqrt{3}) - (-1-i\sqrt{3})}{2} = i\sqrt{3} = z$

So $\omega^2 - \omega = -z$, and the determinant is $3(-z) = -3z$.

Therefore $3k = -3z \implies k = -z$.

Answer: (2) $-z$

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PYQ 2 (AIEEE 2005)

If the cube roots of unity are $1, \omega, \omega^2$, then the roots of the equation $(x-1)^3 + 8 = 0$ are: (a) $-1, 1+2\omega, 1+2\omega^2$ (b) $-1, 1-2\omega, 1-2\omega^2$ (c) $-1, -1, -1$ (d) $-1, -1+2\omega, -1-2\omega^2$

Solution:

$(x-1)^3 = -8 = 8 \cdot (-1) = 8e^{i\pi}$

Taking cube roots: $x - 1 = 2 \cdot \sqrt[3]{-1}$

The cube roots of $-1$ are: $-1, -\omega, -\omega^2$

So:

• $x - 1 = 2(-1) = -2 \implies x = -1$
• $x - 1 = 2(-\omega) = -2\omega \implies x = 1 - 2\omega$
• $x - 1 = 2(-\omega^2) = -2\omega^2 \implies x = 1 - 2\omega^2$

Answer: (b) $-1, 1-2\omega, 1-2\omega^2$

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PYQ 3 (JEE Main 2018)

If $\alpha, \beta \in \mathbb{C}$ are the distinct roots of the equation $x^2 - x + 1 = 0$, then $\alpha^{101} + \beta^{107}$ is equal to: (a) $-1$ (b) $0$ (c) $1$ (d) $2$

Solution:

$x^2 - x + 1 = 0$ has roots that are related to cube roots of unity.

Note: $x^3 + 1 = (x+1)(x^2 - x + 1)$

So if $\alpha$ is a root of $x^2 - x + 1 = 0$, then $\alpha^3 = -1$.

The roots are $\alpha = -\omega$ and $\beta = -\omega^2$ (or vice versa).

Calculate: $\alpha^{101} = (-\omega)^{101} = (-1)^{101} \cdot \omega^{101} = -\omega^{101}$

Since $\omega^3 = 1$: $\omega^{101} = \omega^{99+2} = \omega^2$

So $\alpha^{101} = -\omega^2$

Similarly: $\beta^{107} = (-\omega^2)^{107} = (-1)^{107} \cdot \omega^{214} = -\omega^{214}$

$\omega^{214} = \omega^{213+1} = \omega^1 = \omega$

So $\beta^{107} = -\omega$

Therefore: $\alpha^{101} + \beta^{107} = -\omega^2 - \omega = -(\omega + \omega^2) = -(-1) = 1$

Answer: (c) $1$

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PYQ 4 (IIT JEE Pattern)

If $\omega (\neq 1)$ is a cube root of unity and $(1 + \omega)^7 = A + B\omega$, find $(A, B)$.

Solution:

First, simplify $1 + \omega$: Since $1 + \omega + \omega^2 = 0$, we have $1 + \omega = -\omega^2$.

So: $(1 + \omega)^7 = (-\omega^2)^7 = -\omega^{14}$

Since $\omega^3 = 1$: $\omega^{14} = \omega^{12} \cdot \omega^2 = \omega^2$

Therefore: $(1 + \omega)^7 = -\omega^2 = -\left(\frac{-1-i\sqrt{3}}{2}\right) = \frac{1 + i\sqrt{3}}{2}$

But we need to express this as $A + B\omega$.

Using $\omega^2 = -1 - \omega$: $-\omega^2 = 1 + \omega$

So $A + B\omega = 1 + 1 \cdot \omega$

Answer: $(A, B) = (1, 1)$

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PYQ 5 (JEE Main 2021)

If $a$ and $b$ are real numbers such that $(2 + \alpha)^4 = a + b\alpha$ where $\alpha = \frac{-1 + i\sqrt{3}}{2}$, then $a + b$ is equal to:

Solution:

Note that $\alpha = \omega$ (cube root of unity).

$(2 + \omega)^4 = ?$

Let's compute step by step: $(2 + \omega)^2 = 4 + 4\omega + \omega^2 = 4 + 4\omega + \omega^2$

Using $1 + \omega + \omega^2 = 0 \implies \omega^2 = -1 - \omega$: $= 4 + 4\omega - 1 - \omega = 3 + 3\omega$

$(2 + \omega)^4 = (3 + 3\omega)^2 = 9(1 + \omega)^2 = 9(1 + 2\omega + \omega^2)$ $= 9(1 + 2\omega - 1 - \omega) = 9\omega$

So $a + b\omega = 9\omega = 0 + 9\omega$

Thus $a = 0$, $b = 9$, and $a + b = 9$.

Answer: 9

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Power Reduction Technique

To evaluate $\omega^n$ for large $n$:

1. Find $n \mod 3$
2. Use: $\omega^{3k} = 1$, $\omega^{3k+1} = \omega$, $\omega^{3k+2} = \omega^2$

Example: $\omega^{100} = \omega^{99+1} = \omega^{3 \times 33 + 1} = \omega$

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Useful Identities for Quick Computation

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nth Roots of Unity (Extension)

For $n$th roots of unity: $z^n = 1$

The roots are: $e^{i\frac{2\pi k}{n}}$ for $k = 0, 1, 2, \ldots, n-1$

Key Property: Sum of all $n$th roots of unity = $0$ (for $n > 1$)

Example: Fourth roots of unity are $1, i, -1, -i$, and $1 + i + (-1) + (-i) = 0$.

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Summary: Quick Reference Card

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Pro Tips for JEE

1. When you see $\sqrt{3}$ and $\pm 1$: Immediately think of converting to $\omega$ form.

2. For high powers: Reduce using $\omega^3 = 1$.

3. For products like $(1+\omega)(1+\omega^2)$: Use the identity $(1+\omega)(1+\omega^2) = 1$.

4. In determinants with $\omega$: Use $1 + \omega + \omega^2 = 0$ to simplify rows/columns.

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Related Topics: Substitution Hack | Rotation Theorem