De Moivres Theorem for Integral and Rational Powers Explained

De Moivre's Theorem is a fundamental tool in Mathematics that is used to connect two different branches of Mathematics, i.e., trigonometry and complex numbers. De Moivre's Theorem is significant in complex analysis and forms a basis of number theory. In this article, we will discuss the detailed proof of the theorem and also about the law of rational indices.

Table of Contents

• Introduction 

• History of Abraham de Moivre

• Statement of De Moivre's Theorem 

• Proof of De Moivre's Theorem 

• Applications of De Moivre's Theorem

• Limitations of De Moivre's Theorem

History of Abraham De Moivre

Abraham De Moivre

Image Credit: Wikimedia

Name: Abraham de Moivre

Born:  26 May 1667

Died: 27 November 1754

Field: Mathematics

Nationality: French

Statement of De Moivre's Theorem

De Moivre's Theorem states the relationship between complex numbers and trigonometry. De Moivre’s Theorem for integral and rational powers has relation as follows:

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

Proof of De Moivre's Theorem

Here, we will use the method of Mathematical Induction to prove De Moivre's Theorem. 

To Prove: $(\cos x+i \sin x)^{n}=\cos (n x)+i \sin (n x) \ldots(i)$

Firstly, we will see whether the result holds for $n=1$

So, For $n=1$,

we have

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

which is true.

The result holds true for $n=1$.

Now let us assume that result is true for $n=k$.

Then, we have

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

So, Now we will prove that result is also true for $n=k+1$.

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

$=(\cos (k x)+i \sin (k x))(\cos x+i \sin x) \quad[U \operatorname{sing}(i)]$

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

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

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

Hence the result is proved.

Since the theorem is true for $n=1$ and $n=k+1$, it is true $\forall n \geq 1$.

De Moivre's Theorem When Power is in Fraction Form

If $z=r(\cos \theta+i \sin \theta)$ and $n$ is a positive integer

then De Moivre's Theorem states that: $z^{\dfrac{1}{n}}=r^{\dfrac{1}{n}}\left[\cos \left(\dfrac{2 k \pi+\theta}{n}\right)+i \sin \left(\dfrac{2 k \pi+\theta}{n}\right)\right]$

where $k=0,1,2,3, \ldots,(n-1)$.

Applications of De Moivre's Theorem

• De Moivre's Theorem is used to find the nth root of complex numbers.

• De Moivre's Theorem is used in computing and programming.

• De Moivre's Theorem connects complex numbers with trigonometry.

Limitations of De Moivre's Theorem

• De Moivre's theorem gives multiple valued results in the case of non-integers power.

• De Moivre's Theorem does not work for all non-integers powers.

Solved Examples

1: Find the value of $(1-\sqrt{3} i)^{5}$ using the De Moivre formula.

Ans. Let $z=1-\sqrt{3} i=a+i b$

Its modulus is, $r=\sqrt{\left(a^{2}+b^{2}\right)}=\sqrt{(1+3)}=2$.

$\alpha=\tan ^{-1}\left|\dfrac{b}{a}\right|=\tan ^{-1} \sqrt{3}=\dfrac{\pi}{3}$

Since $a>0$ and $b<0, \theta$ is in the $4^{t h}$ quadrant.

So,

$\theta=2 \pi-\dfrac{\pi}{3}=\dfrac{5 \pi}{3}$

Thus,

$z=r(\cos \theta+i \sin \theta)=2\left(\cos \dfrac{5 \pi}{3}+i \sin \dfrac{5 \pi}{3}\right)$

Now,

$z^{5}=\left(2\left(\cos \dfrac{5 \pi}{3}+i \sin \dfrac{5 \pi}{3}\right)\right)^{5}$

By De Moivre formula,

$z^{5}=2^{5}\left(\cos \dfrac{25 \pi}{3}+i \sin \dfrac{25 \pi}{3}\right)$

$=32\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2 i}\right)$

$=16+16 \sqrt{3} i$

2. If $z=(\cos \theta+i \sin \theta)$, show that $z^{n}+\dfrac{1}{z^{n}}=2 \cos n \theta$ and $z^{n}-\left[\dfrac{1}{z^{n}}\right]=2 i \sin n \theta$

Ans. Let $z=(\cos \theta+i \sin \theta)$.

By De Moivre's Theorem,

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

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

$\dfrac{1}{z^{n}}=z^{-n}=\operatorname{cosn} \theta-i \sin \theta$

Therefore,

$z^{n}+\dfrac{1}{z^{n}}=(\operatorname{cosn} \theta+i \sin \theta)+(\operatorname{cosn} \theta-i \sin \theta \theta) $

$z^{n}+\dfrac{1}{z^{n}}=2 \operatorname{cosn} \theta$

Similarly,

$z^{n}-\dfrac{1}{z^{n}}=(\operatorname{cosn} \theta+i \sin n \theta)+(\cos n \theta-i \sin n \theta) $

$z^{n}-\dfrac{1}{z^{n}}=2 i \sin n \theta$

3. Solve $\left(\sin \dfrac{\pi}{6}+i \cos \dfrac{\pi}{6}\right)^{18}$.

Ans. We have,

$\sin \dfrac{\pi}{6}+i \cos \dfrac{\pi}{6}=i\left(\cos \dfrac{\pi}{6}-i \sin \dfrac{\pi}{6}\right)$.

Raising to the power of 18 on both sides it gives,

$\left(\sin \dfrac{\pi}{6}+i \cos \dfrac{\pi}{6}\right)^{18}=(i)^{18}\left(\cos \dfrac{\pi}{6}-i \sin \dfrac{\pi}{6}\right)^{18} $

$=(-1)\left(\cos \dfrac{18 \pi}{6}-i \sin \dfrac{18 \pi}{6}\right) $

$=-(\cos 3 \pi-i \sin 3 \pi)=1+0 i$

Therefore, $\left(\sin \dfrac{\pi}{6}+i \cos \dfrac{\pi}{6}\right)^{18}=1$.

Conclusion

In the article, we have discussed the detailed proof of De Moivre's Theorem for natural and rational roots. De Moivre’s Theorem is an important component of number theory and has a wide range of applications in computing. So, we conclude that De Moivre Theorem, which acts as a connecting link between complex analysis and trigonometry, is an important component of mathematics.

Important Points to Remember

• If $n$ is integer: $(\cos x+i \sin x)^{n}=\cos (n x)+i \sin (n x)$.

• If the power is in fraction form: $z^{\dfrac{1}{n}}=r^{\dfrac{1}{n}}\left[\cos \left(\dfrac{2 k \pi+\theta}{n}\right)+i \sin \left(\dfrac{2 k \pi+\theta}{n}\right)\right]$.

Related Links

• Complex Numbers

• Trigonometric Functions

• Questions and Answers