There is a beautiful theorem attached to our study of Complex numbers, and yes, it has a romantic name to it, like most other theorems in mathematics, known as De Moivre's Theorem. The theorem, though, is not as complicated as it sounds, but is definitely complex (pun intended). De Moivre's Theorem will serve as a gateway to more advanced theory of complex numbers and will help in tackling some crucial problems on the way.
De Moivre's theorem definition
Suppose you have a complex number \(z=1+2i\), and you are asked to calculate the square of this complex number: \(z^2=(1+2i)^2\).
Well, easy enough:
$$ \begin{aligned} z^2 & =(1+2i)(1+2i) \\ &=1+2i+2i-4 \\ z^2 &=-3+4i \end{aligned}$$
But things get tedious as you go ahead, what if you were asked to calculate higher powers of it, like \(z^4, \ z^5,...\) and so on. Binomial expansion can be effectively used, but it becomes impractical as the powers get larger and larger.
For such large powers, there is a very elegant theorem by which one can come up with a concise formula for any integer power of a complex number:
De Moivre's Theorem (also known as De Moivre's identity) is named after the French mathematician Abraham de Moivre, although the theorem never appeared in any of his works.
The statement of the theorem is given as follows:
For a complex number \(z=r(\cos x +i\sin x)\), \(r\) being the modulus of the complex number and for some integer \(n\), raising \(z\) to the power of \(n\), we get:
$$z^n=r^n(\cos {nx}+i \sin {nx})$$
This is a mighty formula in itself since raising the power of a complex number to an integer, you simply have to multiply the argument of the complex by the integer the complex number is raised to.
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 \(n\)
Note that De Moivre's theorem does not hold for rational values of \(n\).
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 \(1\).
Find the cube root of \(1\).
Solution:
Let \(x^3=1\) where you are asked to find the values of \(x\). One root can be observed straight away, which is \(x=1\).
Writing \(1\) in polar form:
$$1=1+0i=s(\cos \alpha +i\sin \alpha)$$
Comparing the real and imaginary part, you have \(s=1\), \(\cos \alpha=1\) and \(\sin \alpha=0\) which gives \(\alpha=0\).
Let the root be \(x=r(\cos \theta+i \sin \theta)\) so that,
$$r=s^{1/n}=1 \ \ \text{and} \ \ \theta=\frac{\alpha +2 \pi k}{n}=\frac{2 \pi k}{3}$$
Thus, you get the solution set as:
$$x=\cos \left( {\frac{2 \pi k}{3}} \right) +i \sin \left( {\frac{2 \pi k}{3}} \right) $$
All now left is to substitute \(k=0, \ 1, \ 2\) to have the different roots.
For \(k=0\), \(x_1=1\), unsurprisingly, since it was the most obvious one.
For \(k=1\),
$$ \begin{aligned} x_2 & =\cos \left( \frac{2 \pi}{3} \right)+i\sin \left( \frac{2 \pi}{3} \right) \\ & = -\frac{1}{2}+i\frac{\sqrt{3}}{2} \\ \therefore \ x_2 & =\frac{-1+i\sqrt{3}}{2} \end{aligned}$$
For \(k=2\),
$$ \begin{aligned} x_3 & =\cos \left( \frac{4 \pi}{3} \right) + i \sin \left( \frac{4 \pi}{3} \right) \\ & = -\frac{1}{2}+i\frac{-\sqrt{3}}{2} \\ \therefore \ x_3 & =\frac{-1-i\sqrt{3}}{2} \end{aligned}$$
Thus, the cube roots of \(1\) are \(1, \ \displaystyle \frac{-1+i\sqrt{3}}{2}, \ \frac{-1-i\sqrt{3}}{2}\).
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.
Find the value of \((\sin \left( \frac{\pi}{6} \right) + i \cos \left( \frac{\pi}{6} \right))^{18}\).
Solution:
Notice that the complex number is of the form \( \sin x+i \cos x\) and not \(\cos x+i \sin x\), so you need to convert it :
$$\begin{aligned} \left(\sin \frac{\pi}{6}+i \cos \frac{\pi}{6}\right)^{18} &=\left(i\left(\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}\right)\right)^{18} \\ &=i^{18}\left(\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}\right)^{18} \end{aligned}$$
Now you can safely apply de Moivre's identity:
$$\begin{aligned} \left(\sin \frac{\pi}{6}+i \cos \frac{\pi}{6}\right)^{18} &=(-1)(\cos 3 \pi-i \sin 3 \pi) \\ &=-(-1-0 i) \\ &=1+0i \\ \therefore \ \left(\sin \frac{\pi}{6}+i \cos \frac{\pi}{6}\right)^{18} &=1 \end{aligned}$$
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
If \(z=(\cos \theta+i \sin \theta)\), show that \(z^n+\frac{1}{z^n}=2 \cos n \theta\) and
\(z^n-\frac{1}{z^n}=2 i \sin n \theta\), where \(n \in \mathbb{Z}\)
Solution:
Let \(z=(\cos \theta+i \sin \theta)\), taking the \(n^{\text {th }}\) power:
$$z^n=\cos n \theta+i \sin \theta$$
Replacing \(n\) with \(-n\) :
$$\begin{aligned} z^{-n} &=\cos(-n \theta)+i \sin (-n \theta) \\\therefore \ \frac{1}{z^n} &=\cos n \theta-i \sin n \theta \end{aligned}$$Adding \(z^n\) and \(\frac{1}{z^n}\):$$ \begin{aligned} z^n+\frac{1}{z^n} &=\cos n \theta+i \sin \theta+\cos n \theta-i \sin n \theta \\\therefore \ z^n+\frac{1}{z^n}&=2 \cos n \theta \end{aligned}$$
Hence, the first identity has been proved.
Subtracting \(1 / z^n\) from \(z^n\) :
$$\begin{aligned} z^n-\frac{1}{z^n} &=\cos n \theta+i \sin n \theta-\cos n \theta+i \sin n \theta \\ \therefore \ z^n-\frac{1}{z^n}&=2 i \sin n \theta \end{aligned}$$
Thus, the second identity has also been proved.
Expand the complex number \((1-\sqrt{3} i)^5\) using De Moivre's identity.
Solution:
Since the complex number in not in polar form, we need to convert it first. To do so, we have to calculate the modulus of it, let \(z=1-i \sqrt{3}\) :
$$\begin{aligned} r &=\sqrt{a^2+b^2} \\&=\sqrt{1+3} \\ \therefore \ r &=2 \end{aligned}$$
For the argument of this complex number:
$$\begin{aligned} \theta &=\tan ^{-1}\left(\frac{b}{a}\right) \\&=\tan ^{-1}\left(\frac{\sqrt{3}}{1}\right) \\\therefore \ \theta &=\pi / 3 \end{aligned}$$
But be careful, this in not the argument. Notice that the complex number lies in the fourth quadrant, so:
$$\begin{aligned}\therefore \phi &=2 \pi-\theta \\&=2 \pi-\frac{\pi}{3} \\\therefore \ \phi &=\frac{5 \pi}{3} \end{aligned}$$
which is the correct argument of the complex number.
Now, the polar form can be written down as:
$$\begin{aligned} z &=r(\cos \phi+i \sin \phi) \\&=2\left(\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3}\right) \end{aligned}$$
You now have the form on which you can apply De Moivre's identity:
$$\begin{aligned} z^5 &=2^5\left(\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3}\right)^5 \\&=32\left(\cos \frac{25 \pi}{3}+i \sin \frac{25 \pi}{3}\right) \\\therefore \ z^5 &=32\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right) \\&=16+16 \sqrt{3} i \end{aligned}$$
Thus, \((1-i \sqrt{3})^5=16+16 \sqrt{3} i\).
De Moivre's Theorem – Key takeaways
- De Moivre's Theorem states that a complex number \(z=r(\cos x+i\sin x)\), when raised to an integer power of \(n\), it can be written as \(z^n=r^n (\cos {nx} + i\sin {nx})\).
- The theorem can be applied to finding roots of equations, for \(x^n=a+ib\), where \(a+ib)=s(\cos \alpha + i\sin \alpha)\). The solution of the equation \(x^n=a+ib\) is then given by \(x=s^{1/n} (\cos \left( \frac{\alpha + 2 \pi k}{n} \right) +i \sin \left( \frac{\alpha + 2 \pi k}{n} \right)) \).
- The prime application of using De Moivre's Theorem is using it to expand a complex number of an integer power without the use of binomial theorem.
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