Mastering De Moivre's Theorem for Competitive Exams

De Moivre's Theorem is a cornerstone for solving problems involving powers and roots of complex numbers, frequently appearing in competitive exams like JEE. This module will guide you through understanding and applying this powerful theorem.

What are Complex Numbers?

Before diving into De Moivre's Theorem, let's quickly recap complex numbers. A complex number is typically expressed in the form $a + bi$ , where $a$ and $b$ are real numbers, and $i$ is the imaginary unit, defined as $i^2 = -1$ . The number $a$ is called the real part, and $b$ is called the imaginary part. Complex numbers can also be represented in polar form, which is crucial for De Moivre's Theorem.

Polar Form of Complex Numbers

A complex number $z = a + bi$ can be represented in polar form as $z = r(\cos heta + i \sin heta)$ , where:

The polar form is often abbreviated using Euler's formula as $z = re^{i\theta}$ , where $e^{i\theta} = \cos \theta + i \sin \theta$ . This form is particularly elegant for multiplication and exponentiation.

Introducing De Moivre's Theorem

Applications of De Moivre's Theorem

De Moivre's Theorem has several key applications, particularly in competitive mathematics:

Example Problem

Let's find the value of $(1 + i)^8$ .

Step 1: Convert to Polar Form For $z = 1 + i$ , the modulus is $r = \sqrt{1^2 + 1^2} = \sqrt{2}$ . The argument $\theta$ satisfies $\tan \theta = \frac{1}{1} = 1$ . Since $1+i$ is in the first quadrant, $\theta = \frac{\pi}{4}$ . So, $1 + i = \sqrt{2} \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right)$ .

Step 2: Apply De Moivre's Theorem Using De Moivre's Theorem with $n=8$ : $(1 + i)^8 = \left[ \sqrt{2} \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right) \right]^8$ $= (\sqrt{2})^8 \left( \cos\left(8 \times \frac{\pi}{4}\right) + i \sin\left(8 \times \frac{\pi}{4}\right) \right)$ $= 2^4 \left( \cos(2\pi) + i \sin(2\pi) \right)$

Step 3: Simplify Since $\cos(2\pi) = 1$ and $\sin(2\pi) = 0$ , we have: $= 16 (1 + i \cdot 0)$ $= 16$

Remember to always convert to polar form before applying De Moivre's Theorem for powers and roots. The polar form $r(\cos heta + i \sin heta)$ is the key!

What is the formula for De Moivre's Theorem when raising a complex number

z = r(\cos \theta + i \sin \theta)

to the power of

n

$z^n = r^n(\cos(n\theta) + i \sin(n\theta))$

Common Pitfalls and Tips

Practice Makes Perfect

The best way to master De Moivre's Theorem is through consistent practice. Work through a variety of problems involving powers, roots, and trigonometric identities derived from the theorem. Focus on problems from past JEE papers to get accustomed to the exam's style and difficulty.

Learning Resources

De Moivre's Theorem - Wikipedia(wikipedia)

Provides a comprehensive overview of De Moivre's Theorem, its statement, proof, and applications, including its extension to roots of complex numbers.

Complex Numbers and De Moivre's Theorem - Khan Academy(video)

A clear video explanation of De Moivre's Theorem, demonstrating how to use it with examples for powers of complex numbers.

De Moivre's Theorem - Brilliant.org(blog)

An interactive explanation of De Moivre's Theorem, covering its statement, derivation, and practical applications with engaging examples.

Finding the nth Roots of a Complex Number using De Moivre's Theorem(documentation)

Details on how to find the $n$-th roots of a complex number using the extension of De Moivre's Theorem, with clear formulas and examples.

De Moivre's Theorem - Mathematics Stack Exchange(forum)

A collection of questions and answers related to De Moivre's Theorem, offering diverse problem-solving approaches and insights from the community.

Complex Numbers: Polar Form and De Moivre's Theorem - YouTube(video)

A video tutorial that thoroughly explains the polar form of complex numbers and its direct application with De Moivre's Theorem for solving problems.

De Moivre's Theorem - Proof and Examples(tutorial)

A straightforward explanation of De Moivre's Theorem with step-by-step examples, making it accessible for learners.

JEE Mathematics: Complex Numbers - De Moivre's Theorem(video)

A video specifically tailored for JEE aspirants, focusing on De Moivre's Theorem and its application in competitive exam problems.

Advanced Complex Numbers: De Moivre's Theorem and Roots(paper)

A more advanced PDF document from Oxford University that delves into the theoretical underpinnings and applications of De Moivre's Theorem.

De Moivre's Theorem - MathWorld(documentation)

A detailed mathematical explanation of De Moivre's Theorem, including its formula, extensions, and related mathematical concepts.