de-Moivre's Theorem and its proof

de-Moivre's Theorem, named after Abraham de Moivre is stated as the following.

$( \cos x + i \sin x )^n = \cos {nx} + i \sin {nx} , \quad n \in \mathbb{Z}$

Just like we have algebraic formulas for $(a+b)^n$ for easy simplification, we should also have similar relations for $(a+ib)^n$ and this theorem does exactly that. However, the more important and interesting part of the theorem is that it allows us to visualize what happens to the magnitude and the angle of a complex number when we raise it to some power.

There is one limitation that this only works for integer powers. However, before we go for the applications, let us talk about its proof. You can skip this proof at this stage of the course and loop back to it after covering induction and trigonometry, but I will make the proof simpler so that you can digest it here.

Pre-Requisites for the Proof

The proof uses two main areas, mathematical induction and a few trigonometric relations.

Mathematical Inductions:

To prove $( \cos x + i \sin x )^n = \cos {nx} + i \sin {nx}$ , We call this whole thing a statement and denote it by $S(n)$ . We do not know for which values of $n$ is the given statement true. Mathematical induction is the method of proving such statements in two steps. First, we have to determine that if $S(n)$ is true, then it follows that $S(n+1)$ is also true. Secondly, we have to show that $S(n)$ is true for some base case like $S(0)$ or $S(1)$ .

Simply, mathematically induction works like this, we show that if $S(3)$ is correct, then $S(4)$ must also be correct. It follows that than $S(5)$ , $S(6)$ , and so on are all correct, but only if our initial assumption that $S(3)$ is correct. This is step one.

Then we simplify an easy base case and manually prove that $S(1)$ is true, which then from step one confirms that $S(2)$ must also be true and then by extension $S(3)$ is also true. These two steps, when done correctly will prove that $S(n)$ is true for all $n$ as we have defined them.

Chapter 8 will deal with Mathematical Induction in more detail.

Trigonometric Identities

The proof of de-Moivre's Theorem requires already understanding and holding the following trigonometric identities to be true. These identities will be taught in later chapters.

$\cos(-x) = \cos(x)$

$\sin(-x) = -\sin(x)$

$\sin^2 x + \cos^2x = 1$

$\sin (x+y) = \sin x \cos y + \cos x \sin y$

$\cos (x+y) = \cos x \cos y - \sin x \sin y$

Prove That $\\ (\cos x + i \sin x )^n = \cos {nx} + i \sin {nx} , \quad n \in \mathbb{Z}$

How do we state de Moivre's theorem for a positive integer n?

$S(n)$ is given as $(\cos x + i \sin x)^n = \cos nx + i \sin nx$

What is our inductive hypothesis?

We assume that $S(k)$ is true for positive $k$

How do we write the inductive hypothesis mathematically?

$(\cos x + i \sin x)^k = \cos kx + i \sin kx$

Our next step is to;

We need to prove that $S(k+1)$ is also true.

SKIP

Which is written as

How do we break down $(\cos x + i \sin x)^{k+1}$ using exponent rules?

$(\cos x + i \sin x)^{k+1} = (\cos x + i \sin x)^k (\cos x + i \sin x)$

How do we substitute our inductive hypothesis?

$= (\cos kx + i \sin kx)(\cos x + i \sin x)$

How do we expand this product of complex numbers?

$= \cos kx \cos x - \sin kx \sin x + i(\cos kx \sin x + \sin kx \cos x)$

What trigonometric identities can we apply to simplify this expression?

$= \cos(kx + x) + i \sin(kx + x)$

How do we simplify $kx + x$ ?

$= \cos(k+1)x + i \sin(k+1)x$

SKIP

This establishes that

What equation have we just proven?

$(\cos x + i \sin x)^{k+1} = \cos(k+1)x + i \sin(k+1)x \quad [1]$

What does this tell us about the inductive step?

Which means that $S(k+1)$ is always true if $S(k)$ is true

What do we need to establish next?

Now we need to prove that some base case is true. $S(1)$ is simple

How do we verify the base case $S(1)$ ?

$S(1) = (\cos x + i \sin x)^1 = \cos(1x) + i \sin(1x) \quad [2]$

SKIP

If $S(1)$ is true, then $S(2)$ is true and so on.

What case do we need to consider next?

Now we need to establish that $S(n)$ is true where $n$ is a negative integer.

SKIP

We write a negative case as

How do we write the negative power mathematically?

$(\cos x + i \sin x)^{-n}$ where $n$ is positive

SKIP

Simplifying this.

How do we rewrite a negative exponent?

$(\cos x + i \sin x)^{-n} = \dfrac{1}{(\cos x + i \sin x)^n}$

How do we apply our proven formula to the denominator?

$= \dfrac{1}{\cos nx + i \sin nx}$

How do we rationalize this complex fraction?

$= \dfrac{1}{\cos nx + i \sin nx} \cdot \dfrac{\cos nx - i \sin nx}{\cos nx - i \sin nx}$

What does the denominator become after rationalization?

$= \dfrac{\cos nx - i \sin nx}{\cos^2 nx + \sin^2 nx}$

How do we simplify the denominator using a fundamental trigonometric identity?

$= \cos nx - i \sin nx$

How can we rewrite this using the properties of even and odd functions?

$= \cos(-nx) + i \sin(-nx)$

SKIP

Which establishes that

What equation have we proven for negative exponents?

$(\cos x + i \sin x)^{-n} = \cos(-nx) + i \sin(-nx) \quad [3]$

What does this prove about negative cases?

Which establishes that $S(-k)$ also holds true if $S(k)$ is true.

What is our final conclusion?

[1], [2], and [3] fully establish that de Moivre's theorem holds true for all positive and negative integers.

Visualizing the Proof

From the previous lessons, we understand that multiplying two complex numbers involve rotating the first number with the angle of the second one. If we multiply a number by itself, then it follows that we are doubling the angle of that number. This is exactly what we see where $n$ , the exponent, is multiplied with the angle inside cosine and sine.

In this interactive graph, you can move around the blue point that represents the complex number to change its modulus and its polar angle. We would suggest keeping the modulus close to one. There is also a slider which you can control to see the positive and negative exponents of $z$ from $-10$ to $10$ .

You can see how the successive powers fan out into a circle at equal intervals of angles based on the initial angle of $z$ visually proving the de-Moivre's theorem.

Visualizing deMoivre's Theorem

End of Lesson

de-Moivre's Theorem and its proof

Pre-Requisites for the Proof

Mathematical Inductions:

Trigonometric Identities

Prove That (cos⁡x+isin⁡x)n=cos⁡nx+isin⁡nx,n∈Z\\ (\cos x + i \sin x )^n = \cos {nx} + i \sin {nx} , \quad n \in \mathbb{Z}(cosx+isinx)n=cosnx+isinnx,n∈Z

Visualizing the Proof

Visualizing deMoivre's Theorem

Prove That $\\ (\cos x + i \sin x )^n = \cos {nx} + i \sin {nx} , \quad n \in \mathbb{Z}$