De Moivre's Theorem

De Moivre's formula

The formula for De Moivre's Theorem can be written down as follows:

$$z^n=(r(\cos x+i \sin x))^n=r^n(\cos{nx}+i\sin{nx})$$

To prove De Moivre's Formula, all you need is the method of 'Proof by Mathematical Induction':

Proof:

Let \(z=r(\cos x+i\sin x)\), and \(n \in \mathbb{N}\):

Let \(P(n)\) be the statement that \(z^n=r^n(\cos{nx}+i\sin{nx})\).

Proceed by proving \(P(1)\):

$$ \begin{aligned} &\text{LHS}=z \\ & \text{RHS}=r(\cos x+i \sin x) \end{aligned}$$

Thus, \(z=r(\cos x+i \sin x)\) (yes, it is as trivial as it seems), which renders \(P(1)\) to be true.

Now assume that \(P(k)\) is true for \(\forall k \in \mathbb{N}\). So, you have \(z^k=r^k(\cos{kx}+i\sin{kx})\).

The goal is to prove the statement \(P(k+1)\) using the fact that \(P(k)\) is true:

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

Starting with \(\text{LHS}\):

$$\begin{aligned} \text{LHS} &=z^{k+1} \\ &=z^k\ z \\ &=r^k(\cos x +i\sin x)^k\ r(\cos x+i\sin x) \end{aligned}$$

Using the statement of \(P(k)\):

$$z^k=r^k(\cos x +i\sin x)^k=r^k(\cos{kx}+i\sin{kx})$$

Substituting this into \(\text{LHS}\):

$$\begin{aligned} \text{LHS}&=r^{k+1}(\cos{kx}+i\sin{kx}) \ (\cos x+i\sin x) \\ &=r^{k+1}(\cos x\cos{kx} +i\cos{kx} \sin x +i\cos x \sin{kx} -\sin x \sin{kx}) \\ \therefore \ \text{LHS} &=r^{k+1}((\cos x \cos{kx} - \sin x \sin{kx})+i(\sin x \cos{kx} +\cos x \sin{kx}) ) \end{aligned}$$

Now you just have to recognise the solution by recalling the trigonometric identities for the sine and cosine functions:

$$\begin{aligned} &\cos{(a+b)}=\cos a \cos b -\sin a \sin b \\ &\sin{(a+b)}=\sin a \cos b+\cos a \sin b \end{aligned}$$

Substituting these formulae into \(\text{LHS}\):

$$\begin{aligned} \text{LHS} &=r^{k+1} (\cos{(k+1)x}+i\sin{(k+1)x}) \\ &=\text{RHS} \\ \therefore \ \text{LHS} &=\text{RHS}\end{aligned}$$

which is what you were asked to prove.

Hence, assuming \(P(k)\) to be true, it implies that \(P(k+1)\) is true. Thus, by the principle of mathematical induction, \(P(n)\) is true \(\forall n \in \mathbb{N}\).

For all the negative integers, the theorem is yet to be proved. To do so, use the result you proved for the positive integers \(n\):

$$z^n=r^n(\cos{nx}+i\sin{nx})$$

Replacing \(n\) by \(-n\):

$$\begin{aligned} z^{-n} &=r^{-n}(\cos (n x)+i \sin (n x))^{-1} \\ &=r^{-n}\frac{1}{\cos (nx)+i \sin (n x)} \cdot\left(\frac{\cos (n x)-i \sin( n x)}{\cos (n x)+i \sin (nx)}\right) \\ &=r^{-n}\frac{\cos(n x)-i \sin (n x)}{\cos^2 (nx)+\sin^2 (nx)} \\ &=r^{-n} \frac{\cos (-n x)+i(\sin (-n x))}{1} \\ \therefore \ z^{-n} & = r^{-n}(\cos(-n x)+i \sin (-n x)) \end{aligned}$$

which proves the theorem for all the negative integers \(n\)

Solving using De Moivre's theorem

Suppose you are asked to solve the cubic equation \(x^3-1=0\), what would your answer be? It certainly would be \(x=1\), which is correct, but the answer is incomplete. \(x=1\) is the real root of this cubic equation, but what about the other roots? The degree of the polynomial is \(3\); hence it should have three roots in total.

This where the methods you have learned in high school come to an end (kind of). It turns out that the other roots of this equation are not real, but complex.

Let's try to address this generally and then go back to the problem of finding the cube root of \(1\). In general, we are trying to solve the equation \(x^n=a+ib\) where \(n \in \mathbb{N}\). Writing \(a+ib\) in polar form, we get:

$$a+ib=s(\cos \alpha +i\sin \alpha) $$

where \(s\) is the magnitude of \(a+ib\) and \(\alpha\) is the principal argument of it.

Now, equating this to \(x^n\), but firstly, let \(x=r(\cos \theta +i\sin \theta)\), so that we have polar forms on both sides:

$$\begin{aligned} &x^n = s(\cos \alpha +i\sin \alpha) \\ \therefore \ &(r(\cos \theta +i\sin \theta)^n = s(\cos \alpha +i\sin \alpha)\end{aligned}$$

This where you use De Moivre's theorem on the left-hand side:

$$ r^n(\cos {n\theta} +i \sin{n\theta}) = s(\cos \alpha +i\sin \alpha)$$

Notice that \(s\) and \(\alpha\) are known, since \(a+ib\) is known. So \(r\) and \(\theta\) can be calculated by comparing the coefficients on both side as follows:

$$ r=s^{1/n} \ \ \ \text{and} \ \ \ \theta=\frac{\alpha}{n}$$

Now recall an important property of trigonometry, that sine and cosine functions are periodic i.e. \(\sin \theta =\sin{2\pi k+\theta}\) and \(\cos \theta = \cos{2 \pi k +\theta}\). Hence, in this case, \(\theta= \displaystyle \frac{\alpha +2 \pi k}{n}\).

Therefore, you have \( r=s^{1/n}\) and \(\theta= \displaystyle \frac{\alpha +2 \pi k }{n}\), where \(k \in {1, \ 2, \ 3, ... , \ n-1}\).

Let's get back to finding the cube roots of \(1\).

Application of De Moivre's theorem in expansion

Try to think of some sort of application of De Moivre's theorem, just by looking at it. Let's pen the identity down for that:

$$z^n=r^n(\cos x+i\sin x)^n=r^n(\cos{nx}+i\sin{nx})$$

where \(z=r(\cos x+i\sin x)\). The identity essentially expands the complex number to give a simpler form, which is indeed the most important application of De Moivre's Theorem; expanding integer powers of a complex number.

A complex number might be expressed in one of the two forms: \(z=a+ib\) and \(z=r(\cos x+i\sin x)\).

The latter form is straightforward to work with; apply De Moivre's identity directly.

Regarding the former form, convert it into the polar form, on which the identity can then be used. Let's take a look at some of the examples where this is in use.

Binomial theorem and De Moivre's theorem

How would you write \(\sin 3\theta\) in terms of \(\sin \theta\) and \(\cos \theta\)? You shall explore a technique involving binomial theorem and de moivre's theorem that will answer this question. In general, you shall learn how to write \(\sin n\theta\) and \( \cos n\theta\) in terms of \(\sin \theta\) and \(\cos \theta\).

Recall that the Binomial Theorem says that for a polynomial \((a+b)^n\) is expanded as:

$$(a+b)^n=a^n + {n \choose 1} a^{n-1}b+ {n \choose 2} a^{n-2}b^2+...+b^n$$

where \( \displaystyle {n \choose r} = \frac{n!}{r! \ (n-r)!} \).

Using this, you can expand \((\cos \theta +i \sin \theta)^n\) to get:

$$ \begin{aligned} (\cos \theta +i \sin \theta)^n &= \cos^n \theta+i{n \choose 1} \cos^{n-1} \theta \sin \theta+i^2{n \choose 2} \cos^{n-2} \theta \sin^2 \theta + \ ... \ +i^n \sin^n \theta \\ & = \left( \cos^n \theta -{n \choose 2} \cos^{n-2} \theta \sin^2 \theta +{n \choose 4} \cos^{n-4} \theta \sin^4 \theta + \ ... \right) \\ & \ \ \ \ +i\left( {n \choose 1} \cos^{n-1} \theta \sin \theta - {n \choose 3} \cos^{n-3} \theta \sin^3 \theta+ \ ...\right) \end{aligned} $$

Comparing the real part of both sides:

$$ \cos n\theta =\cos^n \theta -{n \choose 2} \cos^{n-2} \theta \sin^2 \theta +{n \choose 4} \cos^{n-4} \theta \sin^4 \theta + \ ...$$

And now comparing the imaginary part:

$$\sin n\theta = {n \choose 1} \cos^{n-1} \theta \sin \theta - {n \choose 3} \cos^{n-3} \theta \sin^3 \theta+ \ ...$$

Thus, you have a formula with which the multiple-angle formulas for sine and cosine functions can be written down.

De Moivre's theorem examples