1. Statement of De Moivre’s Theorem
De Moivre’s Theorem connects complex numbers in polar form with powers and trigonometric functions.
Theorem:
If a complex number is written as
\(z = r(\cos \theta + i \sin \theta)\),
then for any positive integer \(n\),
\(z^n = r^n(\cos n\theta + i \sin n\theta)\).
This formula makes it easy to compute high powers of complex numbers.
2. Idea of the Proof Using Polar Form
The theorem comes naturally from multiplying complex numbers in polar form.
Recall:
\(z_1 z_2 = r_1 r_2 (\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2))\).
If we multiply \(z\) by itself repeatedly:
- \(z^2 = r^2(\cos 2\theta + i \sin 2\theta)\)
- \(z^3 = r^3(\cos 3\theta + i \sin 3\theta)\)
Continuing this pattern gives \(z^n = r^n(\cos n\theta + i \sin n\theta)\).
The full proof is done by mathematical induction, but the geometric intuition is enough for practical applications.
3. Finding Powers of Complex Numbers Using De Moivre’s Theorem
De Moivre’s Theorem is most useful for raising complex numbers to high powers.
3.1. Example 1: Square and Cube of a Complex Number
Let \(z = 2(\cos 30° + i \sin 30°)\).
Square:
\(z^2 = 2^2(\cos 60° + i \sin 60°) = 4(\cos 60° + i \sin 60°)\)
Cube:
\(z^3 = 2^3(\cos 90° + i \sin 90°) = 8(\cos 90° + i \sin 90°)\)
3.2. Example 2: Computing a Higher Power
Find \((1 + i)^8\).
Step 1: Convert to polar form.
- \(r = \sqrt{1^2 + 1^2} = \sqrt{2}\)
- \(\theta = 45° = \frac{\pi}{4}\)
So: \(1 + i = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})\).
Step 2: Apply De Moivre:
\((1 + i)^8 = (\sqrt{2})^8 (\cos (8 \cdot \frac{\pi}{4}) + i \sin (8 \cdot \frac{\pi}{4}))\)
\((\sqrt{2})^8 = 16\)
\(8 \cdot \frac{\pi}{4} = 2\pi\)
So:
\((1 + i)^8 = 16(\cos 2\pi + i \sin 2\pi) = 16\).
4. Deriving Trigonometric Identities Using De Moivre’s Theorem
De Moivre’s Theorem helps derive many useful trigonometric identities.
Start with:
\((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta\).
4.1. Example: Formula for cos 3θ
Expand \((\cos \theta + i \sin \theta)^3\) using algebra.
\((\cos \theta + i \sin \theta)^3 = \cos^3 \theta + 3i \cos^2 \theta \sin \theta - 3 \cos \theta \sin^2 \theta - i \sin^3 \theta\)
Group real and imaginary parts:
Real part gives:
\(\cos 3\theta = \cos^3 \theta - 3 \cos \theta \sin^2 \theta\)
Simplify using \(\sin^2 \theta = 1 - \cos^2 \theta\).
4.2. Example: Formula for sin 3θ
Imaginary part gives:
\(\sin 3\theta = 3 \cos^2 \theta \sin \theta - \sin^3 \theta\).
5. Solving Trigonometric Equations Using De Moivre’s Theorem
Some trigonometric equations become easier when expressed using De Moivre’s Theorem.
5.1. Example: Solve cos 5θ = 0
From De Moivre:
\(\cos 5\theta = 0\).
So:
\(5\theta = \frac{\pi}{2} + n\pi\).
Therefore:
\(\theta = \frac{1}{5}\left(\frac{\pi}{2} + n\pi\right)\).