de Moivre's Theorem Revision Notes for Edexcel A-Level Further Mathematics

1.2.2 de Moivre's Theorem

What is de Moivre's Theorem?

de Moivre's Theorem is a powerful result in complex numbers that links trigonometry and complex numbers.
It allows us to raise complex numbers to powers and find roots of complex numbers using their modulus-argument form (or exponential form).

Statement of de Moivre's Theorem:

For any real number $n$ and a complex number $z = r(\cos \theta + i \sin \theta)$ , de Moivre's Theorem states that:

z^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right)

In other words, to raise a complex number to a power, we:

Raise the modulus $r$ to the power $n$

Multiply the argument $\theta$ by $n$

Express the result in the form $\^n (\cos(n\theta) + i \sin(n\theta))$

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Example: Using de Moivre's Theorem to Find Powers

Find $(1 + i)^5$ using de Moivre's Theorem.

Step 1: Convert $1 + i$ to modulus-argument form.

Modulus:

r = |z| = \sqrt{1^2 + 1^2} = \sqrt{2}

Argument:

\theta = \tan^{-1} \left( \frac{1}{1} \right) = \frac{\pi}{4}

Therefore:

1 + i = \sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})

Step 2: Apply de Moivre's Theorem.

To find $(1 + i)^5$ , we use de Moivre's Theorem:

z^5 = (\sqrt{2})^5 \left( \cos \left( 5 \times \frac{\pi}{4} \right) + i \sin \left( 5 \times \frac{\pi}{4} \right) \right)

= 4\sqrt{2} \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right)

Step 3: Simplify the result.

From trigonometry, we know:

\cos \frac{5\pi}{4} = -\frac{1}{\sqrt{2}} \quad \text{and} \quad \sin \frac{5\pi}{4} = -\frac{1}{\sqrt{2}}

So, the final answer is:

(1 + i)^5 = 4\sqrt{2} \left( -\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) = :success[-8 - 8i]

Finding Roots of Complex Numbers Using de Moivre's Theorem

de Moivre's Theorem can also be used to find the nth roots of a complex number.

The general form for the nth roots of $z = r (\cos \theta + i \sin \theta)$ is:

z^{\frac{1}{n}} = r^{\frac{1}{n}} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right)

for $k = 0, 1, 2, \dots, n-1$

This gives $n$ distinct roots, as each root corresponds to a different value of $k$ .

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Example: Finding the Cube Roots using de Moivre's Theorem

Find the cube roots of $z=8$

Step 1: Write $z=8$ in modulus-argument form.

Since $8$ is a real number, it can be written as:

z = 8 (\cos 0 + i \sin 0)

So the modulus is $8$ and the argument is $0$ .

Step 2: Apply de Moivre's Theorem for cube roots.

The cube roots are given by:

z^{\frac{1}{3}} = 8^{\frac{1}{3}} \left( \cos \frac{0 + 2k\pi}{3} + i \sin \frac{0 + 2k\pi}{3} \right)

z^{\frac{1}{3}} = 2 \left( \cos \frac{2k\pi}{3} + i \sin \frac{2k\pi}{3} \right)

for $k = 0, 1, 2$

Step 3: Find the cube roots for different values of $k$

For $k=0$ :

z_0 = 2 (\cos 0 + i \sin 0) = :success[2]

For $k=1$ :

z_1 = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) = 2 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = :success[-1 + i\sqrt{3}]

For $k=2$ :

z_2 = 2 \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) = 2 \left( -\frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = :success[-1 - i\sqrt{3}]

Thus, the three cube roots of $8$ are $2$ , $-1 + i\sqrt{3}$ , and $-1 - i\sqrt{3}$

infoNote

Key Takeaways:

This theorem is extremely useful in handling complex numbers, especially when working with powers and roots in advanced problems.

de Moivre's Theorem states that $z^n = r^n (\cos(n\theta) + i \sin(n\theta))$ , where $z$ is a complex number in modulus-argument form.
It simplifies raising complex numbers to powers and finding roots of complex numbers.
To find roots, de Moivre's Theorem gives multiple solutions, as complex roots occur in conjugate pairs.

de Moivre's Theorem (Edexcel A-Level Further Mathematics): Revision Notes

1.2.2 de Moivre's Theorem

What is de Moivre's Theorem?

Statement of de Moivre's Theorem:

Example: Using de Moivre's Theorem to Find Powers

Finding Roots of Complex Numbers Using de Moivre's Theorem

Example: Finding the Cube Roots using de Moivre's Theorem

Key Takeaways:

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