Ultimate Guide to Complex Powers and Roots: De Moivre's Theorem

Introduction

In Complex Geometry (Part 2), we learned that multiplying two complex numbers in polar form is as simple as multiplying their moduli and adding their arguments. What happens if you multiply a complex number by itself repeatedly? You would raise the modulus to a power, and you would add the argument to itself repeatedly (i.e., multiply the argument). This intuitive geometric idea forms the basis of De Moivre's Theorem, one of the most powerful theorems in HSC Mathematics Extension 2.

Executive Summary

This guide covers the applications of De Moivre's Theorem and the roots of complex numbers:

• De Moivre's Theorem: Proving and applying $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ for integer powers.
• Trigonometric Identities: Using the theorem to derive formulas for $\cos(n\theta)$ and $\sin(n\theta)$.
• Roots of Unity: Finding the $n$th roots of $z = 1$ and plotting them on the unit circle.
• Roots of Complex Numbers: Solving equations of the form $z^n = w$ and interpreting the results as regular polygons on the complex plane.

What is this about?

Imagine being asked to evaluate $(1 + i)^{10}$. Expanding this algebraically using the binomial theorem would be a massive, error-prone undertaking. But if you convert $1+i$ to polar form first, De Moivre's Theorem allows you to evaluate the 10th power in just two lines of working.

Furthermore, just as $x^2 = 4$ has two roots, the equation $z^3 = 8$ has three roots, and $z^n = 1$ has $n$ roots. De Moivre's Theorem provides a systematic method to find all $n$ complex roots, revealing that they form perfectly symmetrical shapes on the Argand diagram.

Main Content

1. De Moivre's Theorem

De Moivre's Theorem states that for any real number $\theta$ and any integer $n$: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ Using the cis shorthand, this is beautifully simple: $(\text{cis}\theta)^n = \text{cis}(n\theta)$

If the complex number has a modulus $r$, the theorem becomes: $[r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))$

Syllabus Note: You must be able to prove this theorem for positive integers using Mathematical Induction, and for negative integers using the property $z^{-n} = \frac{1}{z^n}$ and realizing the denominator.

2. Deriving Trigonometric Identities

De Moivre's Theorem provides a brilliant bridge between complex algebra and trigonometry. We can derive multiple-angle formulas for $\cos(3\theta)$, $\sin(4\theta)$, etc.

Method:

1. Expand $(\cos\theta + i\sin\theta)^n$ using the Binomial Theorem.
2. Expand $(\cos\theta + i\sin\theta)^n$ using De Moivre's Theorem to get $\cos(n\theta) + i\sin(n\theta)$.
3. Equate the real parts of both expansions to find a formula for $\cos(n\theta)$.
4. Equate the imaginary parts of both expansions to find a formula for $\sin(n\theta)$.

3. The $n$th Roots of Unity

Solving the equation $z^n = 1$ means finding the $n$th roots of unity (unity = 1). Over the real numbers, $x^3 = 1$ only has one solution: $x=1$. Over the complex numbers, it has three!

How to find them:

1. Write $1$ in generalized polar form: $1 = \text{cis}(0 + 2k\pi) = \text{cis}(2k\pi)$.
2. The equation is $z^n = \text{cis}(2k\pi)$.
3. Apply De Moivre's Theorem in reverse (take the $n$th root, which is a power of $\frac{1}{n}$): $z = [\text{cis}(2k\pi)]^{\frac{1}{n}} = \text{cis}\left(\frac{2k\pi}{n}\right)$
4. Substitute $k = 0, 1, 2, \dots, n-1$ to find the $n$ distinct roots.

Geometric Interpretation: The $n$ roots of unity are always evenly spaced around the unit circle (a circle of radius 1 centered at the origin). For example, the 3rd roots form an equilateral triangle, and the 4th roots form a square.

4. Roots of General Complex Numbers

To solve $z^n = w$ (where $w$ is any complex number):

1. Convert $w$ to generalized polar form: $w = r\text{cis}(\theta + 2k\pi)$.
2. Take the $n$th root of the modulus, and divide the argument by $n$: $z_k = r^{\frac{1}{n}} \text{cis}\left(\frac{\theta + 2k\pi}{n}\right)$
3. Substitute $k = 0, 1, \dots, n-1$ to find the $n$ roots.

Geometrically, these roots form a regular polygon inscribed in a circle of radius $r^{\frac{1}{n}}$, centered at the origin.

Simple Worked Example

Question: Evaluate $(1 + i)^8$.

Solution: Step 1: Convert $1+i$ to polar form. $r = \sqrt{1^2 + 1^2} = \sqrt{2}$ The point $(1,1)$ is in Quadrant 1, so $\theta = \tan^{-1}(1) = \frac{\pi}{4}$. $1 + i = \sqrt{2}\text{cis}\left(\frac{\pi}{4}\right)$

Step 2: Apply De Moivre's Theorem. $(1 + i)^8 = \left[\sqrt{2}\text{cis}\left(\frac{\pi}{4}\right)\right]^8$ $= (\sqrt{2})^8 \cdot \text{cis}\left(8 \times \frac{\pi}{4}\right)$ $= (2^{\frac{1}{2}})^8 \cdot \text{cis}(2\pi)$ $= 2^4 \cdot \text{cis}(2\pi)$ $= 16 \cdot (\cos(2\pi) + i\sin(2\pi))$

Since $\cos(2\pi) = 1$ and $\sin(2\pi) = 0$: $= 16(1 + 0i) = 16$

Answer: $(1+i)^8 = 16$.

mini-FAQ page

Q: Do I always have to use $k = 0, 1, \dots, n-1$? A: You need to substitute $n$ consecutive integers for $k$ to get all $n$ roots. While $0, 1, 2$ is easiest, you can also use values like $-1, 0, 1$. In fact, using negative values for $k$ is often better because it keeps your resulting arguments within the principal range $(-\pi, \pi]$ automatically!

Q: If the roots of unity form a polygon, what is their sum? A: The sum of the $n$th roots of unity is always zero. Think about it geometrically: if you have 5 equally strong forces pulling outward from a center point in a perfect pentagon, the net force is zero. This is a very common trick used in exam proofs.

Common mistakes to avoid

• Forgetting the $+2k\pi$: When finding roots, if you only write $z^3 = \text{cis}(\pi)$ and divide by 3 to get $\text{cis}(\frac{\pi}{3})$, you have only found one root. You must add the general rotations $2k\pi$ before dividing by $n$ to find the others.
• Using De Moivre's on Cartesian form: You cannot apply the theorem directly to $(x+iy)^n$. You MUST convert to polar form first.

Practice on Vu's Maths Hub

De Moivre's Theorem is a staple of the final questions in the Extension 2 exam.

• Practice Binomial-De Moivre trigonometric proofs in the HSC Complex Numbers Booklet.
• Explore the geometry of roots of unity with the HSC Polynomials Booklet.

Further Readings

• Finish your complex numbers mastery with our final guide: Describing Lines, Curves, and Regions.
• Explore more HSC math resources and full worked solutions at Vu's Maths Hub.

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