De Moivre's Theorem: The Game Changer for Complex Numbers
Master the theorem that simplifies complex number operations, powers, roots, and trigonometric identities for JEE success.
Why De Moivre's Theorem is a JEE Game Changer
De Moivre's Theorem connects complex numbers with trigonometry, creating powerful shortcuts for JEE problems. Based on analysis of past papers, this theorem appears in 3-5 questions every year across both JEE Main and Advanced.
- Solves complex powers in 2 steps instead of 10+
- Finds all roots of complex numbers systematically
- Derives trigonometric identities without memorization
- Saves 5-7 minutes per complex numbers problem
De Moivre's Theorem Statement
For any real number $n$ and integer $n$,
$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$
For a complex number $z = r(\cos\theta + i\sin\theta)$:
$$z^n = r^n[\cos(n\theta) + i\sin(n\theta)]$$
Key Insight:
When raising a complex number to power $n$, the modulus gets raised to power $n$ and the argument gets multiplied by $n$.
This converts complicated binomial expansions into simple trigonometric calculations.
Finding Powers of Complex Numbers
Simplify $(1 + i)^{10}$ using De Moivre's Theorem
Step-by-Step Solution:
Step 1: Convert to polar form: $1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$
Step 2: Apply De Moivre's Theorem:
$(1 + i)^{10} = (\sqrt{2})^{10}\left[\cos\left(10\cdot\frac{\pi}{4}\right) + i\sin\left(10\cdot\frac{\pi}{4}\right)\right]$
Step 3: Simplify: $= 32\left[\cos\left(\frac{5\pi}{2}\right) + i\sin\left(\frac{5\pi}{2}\right)\right]$
Step 4: Reduce angles: $\frac{5\pi}{2} = 2\pi + \frac{\pi}{2}$
Step 5: Final answer: $= 32(0 + i\cdot 1) = 32i$
Time Saved: Direct method requires binomial expansion of $(1+i)^{10}$, which has 11 terms!
Finding Roots of Complex Numbers
Find all cube roots of $8$ using De Moivre's Theorem
Step-by-Step Solution:
Step 1: Write in polar form: $8 = 8(\cos 0 + i\sin 0)$
Step 2: Apply De Moivre's for roots:
$z_k = \sqrt[3]{8}\left[\cos\left(\frac{0 + 2\pi k}{3}\right) + i\sin\left(\frac{0 + 2\pi k}{3}\right)\right]$
Step 3: For $k = 0,1,2$:
• $k=0$: $2(\cos 0 + i\sin 0) = 2$
• $k=1$: $2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = -1 + i\sqrt{3}$
• $k=2$: $2\left(\cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}\right) = -1 - i\sqrt{3}$
Deriving Trigonometric Identities
Express $\cos 3\theta$ in terms of $\cos\theta$ using De Moivre's Theorem
Step-by-Step Solution:
Step 1: By De Moivre: $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$
Step 2: Expand LHS using binomial theorem:
$= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$
Step 3: Equate real parts:
$\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$
Step 4: Use $\sin^2\theta = 1 - \cos^2\theta$:
$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$
🚀 JEE Problem-Solving Strategies
When to Use De Moivre's Theorem:
- Complex numbers raised to high powers
- Finding all roots of complex numbers
- Deriving trigonometric identities
- Solving equations involving complex numbers
Common Mistakes to Avoid:
- Forgetting to convert to polar form first
- Missing roots when finding nth roots
- Incorrect angle reduction
- Not considering all values of k (0 to n-1)
Applications 4-5 Available in Full Version
Includes solving complex equations and advanced trigonometric applications with JEE-level problems
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Simplify $(1 - i)^8$ using De Moivre's Theorem
2. Find all fourth roots of $-16$
3. Express $\sin 4\theta$ in terms of $\sin\theta$ and $\cos\theta$
📋 De Moivre's Theorem Formula Sheet
Main Formulas:
- $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
- $z^n = r^n(\cos n\theta + i\sin n\theta)$
- $z^{1/n} = r^{1/n}\left[\cos\left(\frac{\theta+2\pi k}{n}\right) + i\sin\left(\frac{\theta+2\pi k}{n}\right)\right]$
Key Results:
- $\cos n\theta = \Re[(\cos\theta + i\sin\theta)^n]$
- $\sin n\theta = \Im[(\cos\theta + i\sin\theta)^n]$
- $k = 0, 1, 2, ..., n-1$ for nth roots
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