De Moivre's Theorem - Mathematics.lu

This is the main theorem in the complex numbers unit. Our motivation is the following:
How many solutions are there to the equation $x^2=4$ ? We have two solutions $x=-2$ and $x=2$ .

What about $x^3=8$ ? If the square root gives you two solutions, why does the cubic root only give us 1? The reality is that it gives us 3 solutions, we are used to always think of the real solution, but there are two more solutions that are complex numbers.

Motivational Example: Verify that $z = 2$ , $z = -1 - i\sqrt{3}$ , and $z = -1 + i\sqrt{3}$ are all valid solutions to the cubic equation $z^3 = 8$ by directly cubing each expression.

To prove these are solutions, we will expand each number to the third power using basic algebra and distributive laws, rather than solving for them from scratch.

1. Verification of $z = 2$
This is the immediate real solution:

$z^3 = (2)^3 = 2 \cdot 2 \cdot 2 = 8$

Thus, $z = 2$ is a valid solution.

2. Verification of $z = -1 - i\sqrt{3}$
We first calculate the square of the expression, $(z^2)$ :

$z^2 = (-1 - i\sqrt{3})(-1 - i\sqrt{3})$

$z^2 = (-1)(-1) + (-1)(-i\sqrt{3}) + (-i\sqrt{3})(-1) + (-i\sqrt{3})(-i\sqrt{3})$

$z^2 = 1 + i\sqrt{3} + i\sqrt{3} + i^2(3)$

Substitute $i^2 = -1$ into the equation:

$z^2 = 1 + 2i\sqrt{3} - 3 = -2 + 2i\sqrt{3}$

Now, we multiply this result by the linear term to find the cube, $(z^3 = z^2 \cdot z)$ :

$z^3 = (-2 + 2i\sqrt{3})(-1 - i\sqrt{3})$

$z^3 = (-2)(-1) + (-2)(-i\sqrt{3}) + (2i\sqrt{3})(-1) + (2i\sqrt{3})(-i\sqrt{3})$

$z^3 = 2 + 2i\sqrt{3} - 2i\sqrt{3} - 2i^2(3)$

Notice the imaginary middle terms cancel out perfectly. Substitute $i^2 = -1$ to finish:

$z^3 = 2 - 6(-1) = 2 + 6 = 8$

Thus, $z = -1 - i\sqrt{3}$ is a valid solution.

3. Verification of $z = -1 + i\sqrt{3}$
We expand the square of this expression similarly:

$z^2 = (-1 + i\sqrt{3})(-1 + i\sqrt{3})$

$z^2 = (-1)(-1) + (-1)(i\sqrt{3}) + (i\sqrt{3})(-1) + (i\sqrt{3})(i\sqrt{3})$

$z^2 = 1 - i\sqrt{3} - i\sqrt{3} + i^2(3)$

$z^2 = 1 - 2i\sqrt{3} - 3 = -2 - 2i\sqrt{3}$

Now, we multiply this result by the remaining linear term to find the cube:

$z^3 = (-2 - 2i\sqrt{3})(-1 + i\sqrt{3})$

$z^3 = (-2)(-1) + (-2)(i\sqrt{3}) + (-2i\sqrt{3})(-1) + (-2i\sqrt{3})(i\sqrt{3})$

$z^3 = 2 - 2i\sqrt{3} + 2i\sqrt{3} - 2i^2(3)$

The imaginary components cancel out again. Substitute $i^2 = -1$ :

$z^3 = 2 - 6(-1) = 2 + 6 = 8$

Thus, $z = -1 + i\sqrt{3}$ is a valid solution.

Motivation Note: While expanding these by hand works for power 3, imagine trying to find or check solutions for equations like $z^{10} = 1024$ . This tedious polynomial expansion shows why we need a geometric approach using angles and lengths, which leads directly to De Moivre’s Theorem.

The De Moivre theorem helps us in either of the following cases:

Calculating powers/products of complex numbers,
Calculating roots of complex numbers,
Solving equations of the form $z^n=a$ .

Table of Contents

De Moivre Theorem – Exponential Case

Theorem (De Moivre’s Theorem for Powers): If $z = r(\cos\theta + i\sin\theta)$ is a complex number expressed in polar form and $n$ is any integer, then raising the complex number to the power of $n$ multiplies the argument by $n$ and raises the modulus to the power of $n$ :

$z^n = \big[r(\cos\theta + i\sin\theta)\big]^n = r^n(\cos(n\theta) + i\sin(n\theta))$

This theorem demonstrates that raising a complex number to an integer power corresponds geometrically to scaling its distance from the origin by $r^n$ and rotating its angle around the origin by a factor of $n$ .

This helps us make some calculations faster:

Example: Evaluate $(1 + i\sqrt{3})^5$ by applying De Moivre’s Theorem. Express the final result in standard Cartesian form $x + iy$ .

This is definitely preferred to have to open brackets on an expression to the fifth power. Just as a reminder, Pascal’s triangle gives the formula: $(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$ .

The key to using De Moivre in this form is always the same, write your number in polar form, then use the theorem.

De Moivre Theorem – Root Case

Theorem (De Moivre’s Theorem for Roots): If $z = r(\cos\theta + i\sin\theta)$ is a non-zero complex number expressed in polar form and $n$ is a positive integer, then $z$ has exactly $n$ distinct complex $n$ -th roots. These roots are given by the formula:

$z_k = \sqrt[n]{r} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right) \right)$

where the different solutions are obtained by replacing $k$ by 0, 1, …, $n-1$ .

Geometrically, all $n$ distinct roots lie perfectly on a circle centred at the origin with a radius of $\sqrt[n]{r}$ . They are equally spaced around the circle, separated from one another by an angle of $\frac{2\pi}{n}$ radians. If you are calculating all the fourth roots, each solution will be at a rotation of 90 degrees from the next/previous one. The solutions always form a regular polygon.

Let us now use this theorem to really find all the cubic roots of 8.

Example: Solve the equation $z^3 = 8$ by applying De Moivre’s Theorem for roots. Express all distinct solutions in standard Cartesian form $x + iy$ .

Some Applications of the De Moivre Theorem

The examples below show how useful De Moivre’s theorem can be for dealing with complex numbers and powers.

Example: Evaluate the product $(1 + i)^4 \cdot (1 - i\sqrt{3})^6$ by applying De Moivre’s Theorem. Express the final answer in standard Cartesian form $x + iy$ .

Example: Prove that $i^i$ is a real number. Find its exact value using Euler’s form.

We have used it above to find powers and products of complex numbers. We can also use it in a more general form to show other properties.

Example: Prove that if $|z| = 1$ , then $z^n + z^{-n} = 2\cos(n\theta)$ for any integer $n$ , where $\theta$ is the argument of $z$ .

Our final example will show how we can use De Moivre’s theorem cleverly without necessarily having to use the formula.

Example: Find all fourth roots of 16.

Exercises

Evaluate each of the following expressions by applying De Moivre’s Theorem. Express your final answers in standard Cartesian form $x + iy$ . For numbers starting in Cartesian form, remember to convert them to polar form first.

1. $(1 + i)^6$
2. $(\sqrt{3} - i)^4$
3. $(-2 + 2i\sqrt{3})^3$
4. $\left[2\left(\cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)\right)\right]^4$
5. $\left[\sqrt{3}\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right)\right]^6$

Find all distinct complex solutions for each of the following equations by applying De Moivre’s Theorem for roots. Express your final answers in standard Cartesian form $x + iy$ .

1. $z^3 = 27$
2. $z^4 = -16$
3. $z^3 = 8i$
4. $z^6 = 1$
5. $z^3 = -64i$
6. $z^4 = 81$

Apply De Moivre’s Theorem and the even/odd trigonometric properties to solve the following multi-part problem.

Let $z$ be a complex number on the unit circle such that $|z| = 1$ and its argument is $\theta$ .

Prove that the subtraction of its reciprocal from its power satisfies the general identity:
$z^n - z^{-n} = 2i\sin(n\theta)$
Using this identity, find the exact value of $z^5 - \frac{1}{z^5}$ given that $z = \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)$ .