Welcome to De Moivre’s Theorem!

Hello there! Today, we are diving into one of the most powerful tools in Further Mathematics: De Moivre’s Theorem. If you have ever felt frustrated trying to expand something like \( (1 + i)^{10} \) using binomial expansion, you are going to love this. This theorem provides a massive shortcut for finding powers and roots of complex numbers.

Don't worry if complex numbers felt a bit "imaginary" before—by the end of these notes, you’ll see how this theorem makes calculating them as simple as a few multiplications and additions!

1. The Prerequisite: Polar Form

Before we use the theorem, we must remember how to write a complex number in Polar Form (also known as Modulus-Argument form).

A complex number \( z = x + iy \) can be written as:
\( z = r(\cos \theta + i \sin \theta) \)

  • \( r \) is the modulus (the distance from the origin).
  • \( \theta \) is the argument (the angle from the positive x-axis).

Quick Review: To get here, use \( r = \sqrt{x^2 + y^2} \) and \( \theta = \tan^{-1}(\frac{y}{x}) \). Always check your Argand Diagram to make sure your angle is in the correct quadrant!

2. What is De Moivre’s Theorem?

In simple terms, De Moivre’s Theorem tells us what happens when we raise a complex number in polar form to a power \( n \).

The Formula:
For any integer \( n \):
\( [r(\cos \theta + i \sin \theta)]^n = r^n (\cos n\theta + i \sin n\theta) \)

What does this actually mean?
To raise a complex number to the power of \( n \), you simply:
1. Raise the modulus (\( r \)) to the power of \( n \).
2. Multiply the argument (\( \theta \)) by \( n \).

Analogy: Imagine you are on a merry-go-round. The modulus is how far you are from the center, and the argument is your position on the circle. If you "double" your trip (\( n=2 \)), you move twice as far from the center (if \( r > 1 \)) and you rotate twice as far around the circle!

Example: Finding a Power

Let's find \( [2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})]^3 \).
Step 1: Raise the modulus to the power: \( 2^3 = 8 \).
Step 2: Multiply the angle: \( 3 \times \frac{\pi}{6} = \frac{\pi}{2} \).
Result: \( 8(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) \).

Key Takeaway: De Moivre’s Theorem turns the "scary" job of repeated multiplication into simple arithmetic on the modulus and the angle.

3. Finding the Roots of Complex Numbers

This is where things get really interesting! We can use De Moivre’s Theorem in reverse to find the \( n \)-th roots of a complex number (like the square root, cube root, etc.).

Did you know? A real number like \( 1 \) only has one "real" cube root (which is \( 1 \)), but in the complex world, every number has exactly \( n \) distinct \( n \)-th roots!

How to find roots step-by-step:

To find the \( n \)-th roots of \( z = r(\cos \theta + i \sin \theta) \):

1. Generalize the angle: Remember that adding \( 2\pi \) (a full circle) brings you back to the same spot. So, rewrite your angle as \( (\theta + 2k\pi) \), where \( k \) is an integer.
2. Apply the fractional power: Use De Moivre with the power \( \frac{1}{n} \).
\( z^{1/n} = r^{1/n} [ \cos(\frac{\theta + 2k\pi}{n}) + i \sin(\frac{\theta + 2k\pi}{n}) ] \)
3. Plug in values for \( k \): To find all \( n \) roots, let \( k = 0, 1, 2, ... (n-1) \).

The "Pizza Slicing" Analogy

When you find the \( n \)-th roots of a number, they all lie on a circle in the Argand diagram. They are perfectly spaced out, just like even slices of a pizza! The "crust" of the pizza has a radius of \( \sqrt[n]{r} \).

Common Mistake: Forgetting to add \( 2k\pi \). If you don't do this, you will only find one root instead of all of them!

Key Takeaway: To find roots, find the first one, then "rotate" around the circle by adding \( \frac{2\pi}{n} \) to the angle for each subsequent root.

4. Roots of Unity

A common exam question involves the Roots of Unity. "Unity" is just a fancy mathematical word for the number 1.

The \( n \)-th roots of unity are the solutions to \( z^n = 1 \).
Since \( 1 \) in polar form is \( 1(\cos 0 + i \sin 0) \), the roots are:
\( z = \cos(\frac{2k\pi}{n}) + i \sin(\frac{2k\pi}{n}) \) for \( k = 0, 1, ..., n-1 \).

Quick Trick: The sum of all the \( n \)-th roots of any complex number is always zero. This is because they are perfectly balanced around the origin on the Argand diagram!

5. Summary and Tips for Success

Quick Review Box

  • Powers: Raise \( r \) to \( n \), multiply \( \theta \) by \( n \).
  • Roots: Take the \( n \)-th root of \( r \), divide the generalized angle \( (\theta + 2k\pi) \) by \( n \).
  • Standard Form: Always ensure your final answer is within the required range for the argument (usually \( -\pi < \theta \leq \pi \)).

Common Mistakes to Avoid:

1. The \( r \) mistake: Students often multiply the modulus by \( n \) instead of raising it to the power of \( n \). Don't let that be you!
2. Radians vs Degrees: Most Further Maths questions use radians. Make sure your calculator is in the right mode.
3. The Minus Sign: De Moivre’s Theorem requires a plus sign in the middle: \( \cos \theta \mathbf{+} i \sin \theta \). If you have a minus sign, use the identities \( \cos(-\theta) = \cos \theta \) and \( \sin(-\theta) = -\sin \theta \) to fix it first.

Don't worry if this seems tricky at first! The more you practice converting between Cartesian (\( x+iy \)) and Polar forms, the more natural De Moivre’s Theorem will feel. It is one of the most elegant parts of Pure Maths—enjoy using your new shortcut!