The Hook: The Power of Rotation!
In Thor movies, when Thor swings Mjolnir (his hammer) in circles, each rotation multiplies the momentum. If he rotates it $n$ times, the effect is multiplied $n$ times!
De Moivre’s theorem works similarly — when you raise a complex number to power $n$, you multiply the angle by $n$ and raise the modulus to power $n$. It’s like spinning Thor’s hammer $n$ times!
Real-world application: Signal processing, quantum mechanics, and solving polynomial equations all use this theorem!
Why this matters for JEE: De Moivre’s theorem appears in 1-2 questions and is essential for finding roots of unity and solving higher-degree equations. Master it for quick solutions!
Prerequisites
Before learning De Moivre’s theorem, ensure you understand:
- Modulus and Argument — Polar form $z = r(\cos\theta + i\sin\theta)$
- Trigonometry — Multiple angle formulas
- Exponent laws — $(a^m)^n = a^{mn}$
- Euler’s formula — $e^{i\theta} = \cos\theta + i\sin\theta$
Interactive Demo: Visualize De Moivre’s Theorem
Explore De Moivre’s theorem in action! Use the De Moivre’s Power slider to raise z to different powers and watch:
- The angle multiplies by n (the argument becomes n times theta)
- The modulus raises to power n (r becomes r^n)
Click Animate z^n to see the smooth rotation! Try the Roots of Unity presets to see:
- Cube roots forming an equilateral triangle
- Fourth roots forming a square
- Fifth roots forming a regular pentagon
- How all nth roots sum to zero!
De Moivre’s Theorem: The Statement
For Positive Integer Powers
For any real number $\theta$ and positive integer $n$:
$$\boxed{(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)}$$In simple terms: When you raise a complex number on the unit circle to power $n$, multiply the angle by $n$!
General Form
For any complex number $z = r(\cos\theta + i\sin\theta)$ and positive integer $n$:
$$\boxed{z^n = r^n(\cos(n\theta) + i\sin(n\theta))}$$Breaking it down:
- Modulus: $|z^n| = |z|^n = r^n$ (raise modulus to power $n$)
- Argument: $\arg(z^n) = n\theta$ (multiply angle by $n$)
Using Exponential Form
Using Euler’s formula $e^{i\theta} = \cos\theta + i\sin\theta$:
$$\boxed{(re^{i\theta})^n = r^n e^{in\theta}}$$This is the cleanest form!
Why Does It Work?
Proof for Positive Integers (by Induction)
Base case: $n = 1$
$$(\cos\theta + i\sin\theta)^1 = \cos(1 \cdot \theta) + i\sin(1 \cdot \theta)$$✓
Inductive step: Assume true for $n = k$:
$$(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$$For $n = k + 1$:
$$(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta)$$ $$= [\cos(k\theta) + i\sin(k\theta)][\cos\theta + i\sin\theta]$$Using the multiplication formula for polar forms:
$$= \cos(k\theta + \theta) + i\sin(k\theta + \theta)$$ $$= \cos[(k+1)\theta] + i\sin[(k+1)\theta]$$✓
By induction, theorem holds for all positive integers $n$!
Applications: Finding Powers
Example 1: Simple Power
Find $(1 + i)^{10}$.
Solution:
Step 1: Convert to polar form.
$|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$
$\arg(1 + i) = \frac{\pi}{4}$ (Quadrant I)
So $1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$
Step 2: Apply De Moivre’s theorem.
$$(1 + i)^{10} = (\sqrt{2})^{10}\left[\cos\left(10 \cdot \frac{\pi}{4}\right) + i\sin\left(10 \cdot \frac{\pi}{4}\right)\right]$$ $$= 2^5\left[\cos\frac{10\pi}{4} + i\sin\frac{10\pi}{4}\right]$$ $$= 32\left[\cos\frac{5\pi}{2} + i\sin\frac{5\pi}{2}\right]$$Step 3: Simplify using periodicity.
$\frac{5\pi}{2} = 2\pi + \frac{\pi}{2}$
$$\cos\frac{5\pi}{2} = \cos\frac{\pi}{2} = 0$$ $$\sin\frac{5\pi}{2} = \sin\frac{\pi}{2} = 1$$Answer: $(1 + i)^{10} = 32(0 + i \cdot 1) = 32i$
Memorize powers of common complex numbers:
- $(1 + i)^2 = 2i$
- $(1 + i)^4 = -4$
- $(1 + i)^8 = 16$
- $(\sqrt{3} + i)^6 = -64$
These appear frequently!
Example 2: Negative Powers
Find $\left(\frac{1 + i\sqrt{3}}{2}\right)^{-5}$.
Solution:
First find the polar form:
$\left|\frac{1 + i\sqrt{3}}{2}\right| = \frac{\sqrt{1 + 3}}{2} = \frac{2}{2} = 1$
$\arg\left(\frac{1 + i\sqrt{3}}{2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$
So $\frac{1 + i\sqrt{3}}{2} = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}$
For negative power:
$$\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)^{-5} = \cos\left(-5 \cdot \frac{\pi}{3}\right) + i\sin\left(-5 \cdot \frac{\pi}{3}\right)$$ $$= \cos\left(-\frac{5\pi}{3}\right) + i\sin\left(-\frac{5\pi}{3}\right)$$Converting: $-\frac{5\pi}{3} = -2\pi + \frac{\pi}{3}$
$$= \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$Answer: $\frac{1 + i\sqrt{3}}{2}$
Finding nth Roots of Complex Numbers
The Problem
Given a complex number $w$ and positive integer $n$, find all complex numbers $z$ such that:
$$z^n = w$$These are called the nth roots of $w$.
The Formula
If $w = r(\cos\alpha + i\sin\alpha)$, then the $n$ distinct nth roots are:
$$\boxed{z_k = r^{1/n}\left[\cos\left(\frac{\alpha + 2k\pi}{n}\right) + i\sin\left(\frac{\alpha + 2k\pi}{n}\right)\right]}$$where $k = 0, 1, 2, ..., n-1$.
In simple terms:
- Modulus: Take the $n$th root of modulus: $r^{1/n}$
- Argument: Divide angle by $n$ and add $\frac{2k\pi}{n}$ for each root
Why $n$ Roots?
Every non-zero complex number has exactly $n$ distinct nth roots.
They are equally spaced around a circle of radius $r^{1/n}$ at angular intervals of $\frac{2\pi}{n}$.
Geometric Interpretation
The $n$ nth roots form the vertices of a regular $n$-sided polygon inscribed in a circle of radius $r^{1/n}$!
Examples: Finding Roots
Example 1: Square Roots
Find all square roots of $i$.
Solution:
We need $z^2 = i$.
Step 1: Express $i$ in polar form.
$|i| = 1$, $\arg(i) = \frac{\pi}{2}$
So $i = 1 \cdot \left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)$
Step 2: Apply root formula with $n = 2$.
$$z_k = 1^{1/2}\left[\cos\left(\frac{\frac{\pi}{2} + 2k\pi}{2}\right) + i\sin\left(\frac{\frac{\pi}{2} + 2k\pi}{2}\right)\right]$$For $k = 0$:
$$z_0 = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} = \frac{1 + i}{\sqrt{2}}$$For $k = 1$:
$$z_1 = \cos\left(\frac{\pi}{4} + \pi\right) + i\sin\left(\frac{\pi}{4} + \pi\right) = \cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}$$ $$= -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} = -\frac{1 + i}{\sqrt{2}}$$Answer: $z = \pm\frac{1 + i}{\sqrt{2}}$
Example 2: Cube Roots
Find all cube roots of $8$.
Solution:
We need $z^3 = 8$.
$8 = 8(\cos 0 + i\sin 0)$
$$z_k = 8^{1/3}\left[\cos\left(\frac{0 + 2k\pi}{3}\right) + i\sin\left(\frac{0 + 2k\pi}{3}\right)\right] = 2\left[\cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3}\right]$$For $k = 0$: $z_0 = 2(\cos 0 + i\sin 0) = 2$
For $k = 1$: $z_1 = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3}$
For $k = 2$: $z_2 = 2\left(\cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}\right) = 2\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -1 - i\sqrt{3}$
Answer: $z = 2, -1 + i\sqrt{3}, -1 - i\sqrt{3}$
Example 3: Fourth Roots
Find all fourth roots of $-16$.
Solution:
$-16 = 16(\cos\pi + i\sin\pi)$
$$z_k = 16^{1/4}\left[\cos\left(\frac{\pi + 2k\pi}{4}\right) + i\sin\left(\frac{\pi + 2k\pi}{4}\right)\right]$$ $$= 2\left[\cos\frac{(2k+1)\pi}{4} + i\sin\frac{(2k+1)\pi}{4}\right]$$For $k = 0$: $z_0 = 2\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) = 2 \cdot \frac{1 + i}{\sqrt{2}} = \sqrt{2}(1 + i)$
For $k = 1$: $z_1 = 2\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right) = \sqrt{2}(-1 + i)$
For $k = 2$: $z_2 = 2\left(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right) = \sqrt{2}(-1 - i)$
For $k = 3$: $z_3 = 2\left(\cos\frac{7\pi}{4} + i\sin\frac{7\pi}{4}\right) = \sqrt{2}(1 - i)$
Answer: $\pm\sqrt{2}(1 + i), \pm\sqrt{2}(1 - i)$
Roots of Unity
Definition
The nth roots of unity are the solutions to:
$$z^n = 1$$These are denoted as $1, \omega, \omega^2, ..., \omega^{n-1}$ where $\omega = e^{2\pi i/n}$.
General Formula
$$\boxed{\omega_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n} = e^{2\pi ik/n}}$$where $k = 0, 1, 2, ..., n-1$.
Properties
Sum of roots: $1 + \omega + \omega^2 + ... + \omega^{n-1} = 0$
Product of roots: $1 \cdot \omega \cdot \omega^2 \cdots \omega^{n-1} = (-1)^{n-1}$
Geometric progression: The roots form a GP with first term $1$ and common ratio $\omega$
Regular polygon: The roots are vertices of a regular $n$-gon on the unit circle
Common Mistakes to Avoid
Wrong: The square roots of $4$ are $\pm 2$ (only real roots)
Right: For complex numbers, always find ALL $n$ roots!
For $z^2 = 4 = 4(\cos 0 + i\sin 0)$:
$z_0 = 2(\cos 0 + i\sin 0) = 2$
$z_1 = 2(\cos\pi + i\sin\pi) = -2$
So roots are $\pm 2$ (happens to be real in this case)
After applying De Moivre’s theorem, always reduce the angle to standard range $[0, 2\pi)$ or $(-\pi, \pi]$.
Example: $\cos\frac{10\pi}{4} = \cos\left(2\pi + \frac{\pi}{2}\right) = \cos\frac{\pi}{2} = 0$
Wrong: For $z^3 = 1$, just use $z = \cos 0 + i\sin 0 = 1$
Right: There are 3 roots! Use $k = 0, 1, 2$:
$z_0 = 1$, $z_1 = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}$, $z_2 = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}$
For nth root, the modulus is $r^{1/n}$, NOT $\frac{r}{n}$!
Example: For cube roots of $8$, modulus is $8^{1/3} = 2$, not $8/3$
When to Use De Moivre’s Theorem
Use De Moivre’s theorem when:
- Finding powers of complex numbers (especially high powers)
- Finding nth roots of complex numbers
- Deriving trigonometric identities (like $\cos 3\theta, \sin 5\theta$)
- Solving equations like $z^n = w$
- Working with roots of unity
Key insight: Convert to polar form first, then apply the theorem — much faster than algebraic multiplication!
Practice Problems
Level 1: Foundation (NCERT)
Find $(1 + i)^4$ using De Moivre’s theorem.
Solution:
$1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$
$$(1+i)^4 = (\sqrt{2})^4\left[\cos\left(4 \cdot \frac{\pi}{4}\right) + i\sin\left(4 \cdot \frac{\pi}{4}\right)\right]$$ $$= 4(\cos\pi + i\sin\pi) = 4(-1 + 0) = -4$$Answer: $-4$
Find all square roots of $-1$.
Solution:
$-1 = \cos\pi + i\sin\pi$
$$z_k = \cos\frac{\pi + 2k\pi}{2} + i\sin\frac{\pi + 2k\pi}{2}$$For $k = 0$: $z_0 = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i$
For $k = 1$: $z_1 = \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2} = -i$
Answer: $\pm i$
Find the cube roots of unity.
Solution:
$1 = \cos 0 + i\sin 0$
$$z_k = \cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3}, \quad k = 0, 1, 2$$$z_0 = 1$
$z_1 = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} = \omega$
$z_2 = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} = \omega^2$
Answer: $1, \omega, \omega^2$ where $\omega = \frac{-1 + i\sqrt{3}}{2}$
Level 2: JEE Main
Find $(\sqrt{3} + i)^{10}$.
Solution:
$|\sqrt{3} + i| = \sqrt{3 + 1} = 2$
$\arg(\sqrt{3} + i) = \tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}$
So $\sqrt{3} + i = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$
$$(\sqrt{3} + i)^{10} = 2^{10}\left[\cos\frac{10\pi}{6} + i\sin\frac{10\pi}{6}\right]$$ $$= 1024\left[\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right]$$ $$= 1024\left[\frac{1}{2} - i\frac{\sqrt{3}}{2}\right]$$ $$= 512 - 512i\sqrt{3}$$Answer: $512(1 - i\sqrt{3})$
If $z = 1 + \cos\theta + i\sin\theta$, prove that $z^n = 2^n\cos^n\frac{\theta}{2} \cdot e^{in\theta/2}$.
Solution:
First simplify $z$:
$$z = 1 + \cos\theta + i\sin\theta = (1 + \cos\theta) + i\sin\theta$$Using: $1 + \cos\theta = 2\cos^2\frac{\theta}{2}$ and $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$
$$z = 2\cos^2\frac{\theta}{2} + 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}$$ $$= 2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right)$$ $$= 2\cos\frac{\theta}{2} \cdot e^{i\theta/2}$$Therefore:
$$z^n = \left[2\cos\frac{\theta}{2} \cdot e^{i\theta/2}\right]^n = 2^n\cos^n\frac{\theta}{2} \cdot e^{in\theta/2}$$Proved!
Find all values of $z$ satisfying $z^4 = -1 + i\sqrt{3}$.
Solution:
First, express $-1 + i\sqrt{3}$ in polar form:
$r = \sqrt{1 + 3} = 2$
$\theta = \pi - \tan^{-1}(\sqrt{3}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ (Quadrant II)
So $-1 + i\sqrt{3} = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$
Fourth roots:
$$z_k = 2^{1/4}\left[\cos\frac{\frac{2\pi}{3} + 2k\pi}{4} + i\sin\frac{\frac{2\pi}{3} + 2k\pi}{4}\right]$$ $$= \sqrt[4]{2}\left[\cos\frac{2\pi(3k+1)}{12} + i\sin\frac{2\pi(3k+1)}{12}\right]$$For $k = 0, 1, 2, 3$:
$z_0 = \sqrt[4]{2}\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$
$z_1 = \sqrt[4]{2}\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$
$z_2 = \sqrt[4]{2}\left(\cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6}\right)$
$z_3 = \sqrt[4]{2}\left(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right)$
Answer: Four roots at angles $\frac{\pi}{6}, \frac{2\pi}{3}, \frac{7\pi}{6}, \frac{5\pi}{3}$ with modulus $\sqrt[4]{2}$
Level 3: JEE Advanced
If $\alpha$ is a non-real 5th root of unity, find the value of $\alpha + \alpha^2 + \alpha^3 + \alpha^4$.
Solution:
The 5th roots of unity satisfy $z^5 = 1$, which gives:
$z^5 - 1 = 0$
$(z - 1)(z^4 + z^3 + z^2 + z + 1) = 0$
Since $\alpha \neq 1$ (non-real), it satisfies:
$$\alpha^4 + \alpha^3 + \alpha^2 + \alpha + 1 = 0$$Therefore:
$$\alpha + \alpha^2 + \alpha^3 + \alpha^4 = -1$$Answer: $-1$
Prove that the $n$th roots of unity form a geometric progression and their sum is zero.
Solution:
The nth roots are: $1, \omega, \omega^2, ..., \omega^{n-1}$ where $\omega = e^{2\pi i/n}$
GP property: Each term is $\omega$ times the previous term, so they form a GP with first term $a = 1$ and common ratio $r = \omega$.
Sum: Using GP sum formula:
$$S = \frac{a(r^n - 1)}{r - 1} = \frac{1(\omega^n - 1)}{\omega - 1}$$Since $\omega^n = e^{2\pi i} = 1$:
$$S = \frac{1 - 1}{\omega - 1} = 0$$Proved!
Using De Moivre’s theorem, prove that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.
Solution:
By De Moivre’s theorem:
$$(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$$Expanding LHS using binomial theorem:
$$(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3$$ $$= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$$ $$= (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta)$$Comparing real parts:
$$\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$$ $$= \cos^3\theta - 3\cos\theta(1 - \cos^2\theta)$$ $$= \cos^3\theta - 3\cos\theta + 3\cos^3\theta$$ $$= 4\cos^3\theta - 3\cos\theta$$Proved!
Quick Revision Box
| Concept | Formula | Application |
|---|---|---|
| De Moivre’s Theorem | $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ | Powers |
| General form | $[r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))$ | Powers |
| Exponential form | $(re^{i\theta})^n = r^n e^{in\theta}$ | Quickest! |
| nth roots formula | $z_k = r^{1/n}\left[\cos\frac{\alpha + 2k\pi}{n} + i\sin\frac{\alpha + 2k\pi}{n}\right]$ | $k = 0, 1, ..., n-1$ |
| Roots of unity | $\omega_k = e^{2\pi ik/n}$ | $k = 0, 1, ..., n-1$ |
| Sum of nth roots of unity | $1 + \omega + \omega^2 + ... + \omega^{n-1} = 0$ | Important property |
Related Topics
Within Complex Numbers Chapter
- Modulus and Argument — Foundation for polar form
- Cube Roots of Unity — Special case of nth roots
- Quadratic Equations — Finding complex roots
Math Connections
- Trigonometry — Multiple angle formulas
- Binomial Theorem — Expanding $(a+b)^n$
- Sequences and Series — Geometric progressions
Physics Applications
- Quantum Mechanics — Wave functions and operators
- AC Circuits — Phasor analysis
- Signal Processing — Fourier transforms
Teacher’s Summary
- De Moivre’s Theorem: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ makes powers trivial
- For powers: Raise modulus to power $n$, multiply angle by $n$
- For nth roots: Take nth root of modulus, divide angle by $n$, add $\frac{2k\pi}{n}$ for $k = 0, 1, ..., n-1$
- Every non-zero complex number has exactly $n$ distinct nth roots equally spaced on a circle
- Roots of unity sum to zero and form vertices of regular polygon
“De Moivre’s theorem is your superpower for complex number calculations — memorize it, master it, ace JEE!”
Exam Strategy: For any power or root problem, immediately convert to polar form and use De Moivre. Don’t waste time with algebraic multiplication!