AOS3 Topic 3: De Moivre

De Moivre’s theorem is one of the fundamental theorem of complex numbers which is used to solve various problems of complex numbers. This theorem is also widely used for solving trigonometric functions of multiple angles. De Moivre’s Theorem is also called “De Moivre’s Identity” and “De Moivre’s Formula”. This theorem gets its name from the name of its founder the famous mathematician De Moivre.

De Moivre’s Theorem

De Moivre’s theorem allows us to readily simplify expressions of the form \( z^n \) when \( z \) is expressed in polar form.

De Moivre’s theorem states:

\( (r \text{ cis } \theta)^{n} = r^{n} \text{ cis }(n\theta) \), where \( n \in \mathbb{Z} \).

Proof of De Moivre’s Theorem

This result is usually proved by mathematical induction, but can be explained by a simple inductive argument.

Let \( z = \text{cis} \, \theta \). Then

• \( z^2 \) \( = \text{cis} \, \theta \times \text{cis} \, \theta = \text{cis}(2\theta) \) by the multiplication rule
• \( z^3 = z^2 \times \text{cis} \, \theta = \text{cis}(3\theta) \)
• \( z^4 = z^3 \times \text{cis} \, \theta = \text{cis}(4\theta) \)

Continuing in this way, we see that \( (\text{cis} \, \theta)^n = \text{cis}(n\theta) \), for each positive integer \( n \).

To obtain the result for negative integers, again let \( z = \text{cis} \, \theta \). Then

• \( z^{-1} = \frac{1}{z} = z = \text{cis}(-\theta) \)

For \( k \in \mathbb{N} \), we have

• \( z^{-k} = (z^{-1})^k = \left(\text{cis}(-\theta)\right)^k = \text{cis}(-k\theta) \) using the result for positive integers

Uses of De Moivre’s Theorem

We can raise any complex number (in either rectangular or polar form) to the \(n\)th power easily using De Moivre’s theorem. When given a complex number in rectangular form, make sure to convert it to polar form first.

We can also solve equations that involve complex number roots using De Moivre’s theorem.

Another important use of De Moivre’s theorem is in obtaining complex roots of polynomial equations.

One of its uses is in obtaining relationships between trigonometric functions of multiple angles (like \(\sin 3x\), \(\cos 7x\)) and powers of trigonometric functions.

Examples:

1. Calculation of Powers

Given \( z = 2(\cos(\pi/3) + i \sin(\pi/3)) \), find \( z^5 \).

Solution: Applying De Moivre's Theorem, we have: \[ z^5 = 2^5 \left( \cos(5 \cdot \pi/3) + i \sin(5 \cdot \pi/3) \right) \] \[ = 32 \left( \cos(\pi) + i \sin(\pi) \right) = -32 \]

2. Roots of Complex Numbers

Find the cube roots of \( z = 8(\cos(\pi/4) + i \sin(\pi/4)) \).

Solution: Using De Moivre's Theorem, the cube roots are: \[ z^{1/3} = 8^{1/3} \left( \cos\left(\frac{\pi}{12}\right) + i \sin\left(\frac{\pi}{12}\right) \right) \] \[ z^{1/3} = 2 \left( \cos\left(\frac{\pi}{12} + \frac{2\pi}{3} k\right) + i \sin\left(\frac{\pi}{12} + \frac{2\pi}{3} k\right) \right) \] where \( k = 0, 1, 2 \).

3. Solution of Polynomial Equations

Solve the equation \( z^3 - 8 = 0 \) for complex roots.

Solution: Expressing \( z \) in polar form, we have \( z = 8(\cos(0) + i \sin(0)) \). Applying De Moivre's Theorem, the cube roots are: \[ z^{1/3} = 8^{1/3} \left( \cos\left(\frac{2\pi}{3} k\right) + i \sin\left(\frac{2\pi}{3} k\right) \right) \] \[ z^{1/3} = 2 \left( \cos\left(\frac{2\pi}{3} k\right) + i \sin\left(\frac{2\pi}{3} k\right) \right) \] where \( k = 0, 1, 2 \).

4. Trigonometric Identities

Use De Moivre's Theorem to prove the identity \( \sin(3x) = 3 \sin(x) - 4 \sin^3(x) \).

Solution: Let \( z = \cos(x) + i \sin(x) \). Then \( z^3 = \cos(3x) + i \sin(3x) \). By De Moivre's Theorem, we have: \[ z^3 = (\cos^3(x) - 3 \sin^2(x) \cos(x)) + i (3 \sin(x) \cos^2(x) - \sin^3(x)) \] Comparing the real and imaginary parts, we get the desired identity.

Using De Moivre’s theorem to solve equations

Equations of the form \( z^n = a \), where \( a \in \mathbb{C} \), are often solved by using De Moivre’s theorem.

Write both \( z \) and \( a \) in polar form, as \( z = r \text{ cis } \theta \) and \( a = q \text{ cis } \phi \).

Then \( z^n = a \) becomes

\( (r \text{ cis } \theta)^n = q \text{ cis } \phi \)

\( r^n \cdot \text{ cis}(n\theta) = q \text{ cis } \phi \) (using De Moivre’s theorem)

Compare modulus and argument:

\( r^n = q \)

\( \text{cis}(n\theta) = \text{cis } \phi \)

\( r = \sqrt[n]{q} \)

\( n\theta = \phi + 2k\pi \) where \( k \in \mathbb{Z} \)

\( \theta = \frac{1}{n} (\phi + 2k\pi) \) where \( k \in \mathbb{Z} \)

This will provide all the solutions of the equation.