Complex Numbers Reading Time: 15 min JEE Main & Advanced

Unlocking the Roots: Cube Roots and nth Roots of Unity

Master the geometric beauty and algebraic properties of roots of unity - essential for JEE Complex Numbers.

2-3

Questions in JEE

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Concept Coverage

Key Properties

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Solved Examples

Why Roots of Unity Matter in JEE

Roots of unity are among the most beautiful concepts in complex numbers. They appear in 2-3 questions every year in JEE Main and Advanced, testing your understanding of:

Geometric interpretation of complex numbers
Algebraic manipulation with complex roots
Properties of omega (ω) and their applications
Problem-solving with symmetric expressions

🎯 JEE Insight

Questions on roots of unity often combine complex numbers with algebra, geometry, and even calculus. Mastering this topic gives you an edge in multiple areas.

🧭 Quick Navigation

Cube Roots nth Roots Properties Applications

1. Cube Roots of Unity

Definition and Derivation

The cube roots of unity are the solutions to the equation:

$$ z^3 = 1 $$

Step-by-step Solution:

Step 1: $z^3 - 1 = 0$

Step 2: Factorize: $(z - 1)(z^2 + z + 1) = 0$

Step 3: Solve $z - 1 = 0$ ⇒ $z = 1$

Step 4: Solve $z^2 + z + 1 = 0$ using quadratic formula:

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

$1$

Real root

$\omega$

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

$\omega^2$

$= \frac{-1 - i\sqrt{3}}{2}$

Geometric Interpretation

The cube roots of unity form an equilateral triangle when plotted on the complex plane:

Unit Circle with vertices at:

$1$ at angle $0^\circ$ (real axis)
$\omega$ at angle $120^\circ$
$\omega^2$ at angle $240^\circ$

Key Insight: Each root is obtained by rotating the previous root by $120^\circ$ about the origin.

2. Key Properties of Cube Roots of Unity

Property 1: Sum of Roots

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

Proof: From $z^2 + z + 1 = 0$, the sum of roots is $-1$ (for $\omega$ and $\omega^2$), plus 1 gives 0.

Property 2: Product of Roots

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

Property 3: Cyclic Nature

$\omega^3 = 1$

$\omega^4 = \omega$

$\omega^5 = \omega^2$

General Rule: $\omega^{3n} = 1$, $\omega^{3n+1} = \omega$, $\omega^{3n+2} = \omega^2$

Property 4: Reciprocal Relations

$\frac{1}{\omega} = \omega^2$

$\frac{1}{\omega^2} = \omega$

Property 5: Conjugate Relations

$\overline{\omega} = \omega^2$

$\overline{\omega^2} = \omega$

3. nth Roots of Unity

General Definition

The nth roots of unity are the solutions to:

$$ z^n = 1 $$

General Solution:

The n roots are given by:

$$ z_k = e^{2\pi i k/n} = \cos\left(\frac{2\pi k}{n}\right) + i\sin\left(\frac{2\pi k}{n}\right) $$

where $k = 0, 1, 2, \dots, n-1$

Geometric Interpretation

The nth roots of unity form a regular n-gon inscribed in the unit circle:

Key Features:

All roots lie on the unit circle $|z| = 1$
They are equally spaced at angles of $\frac{2\pi}{n}$
Roots occur in conjugate pairs (except real roots)
The roots are vertices of a regular n-sided polygon

4. Properties of nth Roots of Unity

Property 1: Sum of All Roots

$$ 1 + \alpha + \alpha^2 + \cdots + \alpha^{n-1} = 0 $$

where $\alpha$ is any primitive nth root of unity ($\alpha^n = 1$ but $\alpha^k \neq 1$ for $0 < k < n$)

Property 2: Product of All Roots

$$ 1 \cdot \alpha \cdot \alpha^2 \cdots \alpha^{n-1} = (-1)^{n-1} $$

Property 3: Cyclic Nature

If $\alpha$ is a primitive nth root, then:

$$ \alpha^{kn+r} = \alpha^r $$

The powers repeat with period n

Property 4: Reciprocal Relations

$$ \frac{1}{\alpha^k} = \alpha^{n-k} $$

5. Applications & Solved Examples

Example 1: Simplification using ω

Simplify $(1 + \omega - \omega^2)^3$

Solution:

Step 1: Use $1 + \omega + \omega^2 = 0$ ⇒ $1 + \omega = -\omega^2$

Step 2: Substitute: $(1 + \omega - \omega^2) = (-\omega^2 - \omega^2) = -2\omega^2$

Step 3: Cube: $(-2\omega^2)^3 = -8\omega^6 = -8(\omega^3)^2 = -8(1)^2 = -8$

Answer: $-8$

Example 2: Fourth Roots of Unity

Find all fourth roots of unity and verify their properties

Solution:

Step 1: Solve $z^4 = 1$ ⇒ $z^4 - 1 = 0$

Step 2: Factorize: $(z^2 - 1)(z^2 + 1) = (z-1)(z+1)(z^2+1) = 0$

Step 3: Roots: $1, -1, i, -i$

Step 4: Verify sum: $1 + (-1) + i + (-i) = 0$ ✓

Step 5: Verify product: $1 \cdot (-1) \cdot i \cdot (-i) = (-1) \cdot (-i^2) = (-1) \cdot (1) = -1 = (-1)^{4-1}$ ✓

Example 3: JEE Advanced Problem

If $\omega$ is a complex cube root of unity, find the value of $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$

Solution:

Step 1: Use $1 + \omega + \omega^2 = 0$

Step 2: $1 - \omega + \omega^2 = (1 + \omega + \omega^2) - 2\omega = 0 - 2\omega = -2\omega$

Step 3: $1 + \omega - \omega^2 = (1 + \omega + \omega^2) - 2\omega^2 = 0 - 2\omega^2 = -2\omega^2$

Step 4: Multiply: $(-2\omega)(-2\omega^2) = 4\omega^3 = 4(1) = 4$

Answer: $4$

📚 Quick Revision Formulas

Cube Roots of Unity

Roots: $1, \omega, \omega^2$
$\omega = \frac{-1 + i\sqrt{3}}{2}$
$\omega^2 = \frac{-1 - i\sqrt{3}}{2}$
$1 + \omega + \omega^2 = 0$
$\omega^3 = 1$
$\frac{1}{\omega} = \omega^2$

nth Roots of Unity

$z_k = e^{2\pi i k/n}$
Sum of all roots = 0
Product = $(-1)^{n-1}$
Form regular n-gon on unit circle
Primitive roots: $\alpha^k$ where gcd(k,n)=1

🎯 Practice Problems

Test your understanding with these JEE-level problems:

1. If $\omega$ is cube root of unity, find $(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8)$

Hint: Use $\omega^3 = 1$ to simplify powers

2. Find all sixth roots of unity and show they form a regular hexagon

Hint: Use the formula for nth roots

3. Prove that $1 + \omega^n + \omega^{2n} = 3$ if n is multiple of 3, and 0 otherwise

Hint: Consider cases based on n mod 3

🎯 JEE Exam Strategy

Common Question Types

Simplification using ω properties
Finding specific roots of unity
Geometric interpretation problems
Complex number equations involving roots
Application in solving polynomials

Time-Saving Tips

Memorize the key properties of ω
Use geometric intuition for complex plane
Practice mental calculation with powers of ω
Look for symmetric expressions
Verify answers using multiple methods

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