De Moivre’s Theorem

De Moivre’s Theorem in complex numbers, states that:

For all \(x \in \mathbb{R}\) and \(n \in \mathbb{Z}\)

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

where, \(i = \sqrt{-1}\) 

For example;

\(i^2 =  {(cos \ \frac{\pi}{2} + i sin \ \frac{\pi}{2})}^2\)

= \((cos \ 2 \times \frac{\pi}{2} + i sin \ 2 \times \frac{\pi}{2})\)

= \((cos \ \pi + i sin \ \pi)\) = \(1\)

History

The credit to find the De Moivre’s formula in its recognizable form goes to Abraham De Moivre himself. Abraham De Moivre, in his 1707 A.D. paper in Philosophical Transactions of the Royal Society of London, deduced a formula from which the recognizable form of De Moivre’s formula can be obtained.

In 1749 Euler proved this formula for any real value of \(n\) using Euler’s identity.

Derivation of De Moivre’s Formula from Euler’s Identity

De Moivre’s Formula comes out as a natural consequence of Euler’s Identity. The Euler’s Identity is

 \(e^{ix} = cos \ x + i sin \ x\)

where, \(e\) is the natural logarithmic base and \(x \in \mathbb{R}\).

We know from the exponential law that,

\((e^{a})^b = e ^{ab}\)

Let \(n \in \mathbb{Z}\).

We have, \((e^{ix} )^n = e^{inx}\)

Since \(nx \in \mathbb{R}\)

\(( cos \ x + isin \ x )^n = cos \ nx + isin \ nx\)

De Moivre’s Theorem Proof by Mathematical Induction

Now let’s see proof of the De Moivre’s Theorem by using Principle of mathematical induction.

To Prove: \(( cos \ x + isin \ x )^n = cos \ nx + isin \ nx ; x \in \mathbb{R} \mbox{ & } n \in \mathbb{Z}\)

Proof:

For \(n = 1\) :

\(( cos \ x + isin \ x )^1 = cos \ x + isin \ x = cos \ 1 \times x + isin \ 1 \times x\)

This is trivially true.

For \(n = k\) :

Let the statement be true for some \(n = k ; k \in \mathbb{Z}\).

That is,
\(( cos \ x + isin \ x )^k = cos \ kx + isin \ kx\)

Consider,
\(( cos \ x + isin \ x )^k ( cos \ x + isin \ x )\)
= \(( cos \ kx + isin \ kx ) ( cos \ x + isin \ x )\)
= \(( cos \ kx . cos \ x  – sin \ kx . sin \ x  ) + i ( cos \ kx . sin \ x  + sin \ kx . cos \ x )\)

We know the trigonometric identities:
\(cos ( A + B ) = cos A \ cos B – sin A \ sin B\)
\(sin ( A + B ) = sin A \ cos B + sin B \ cos A\)

Using these,

 \(( cos \ kx . cos \ x  – sin \ kx . sin \ x  ) + i ( cos \ kx . sin \ x  + sin \ kx . cos \ x )\)
= \(cos \ (k+1)x + isin \ (k+1)x\)

Hence proved by induction.

Implications and Applications of De Moivre’s Theorem

Now let’s move on to discussion of implications and applications of this theorem.

Formulae for \(\sin(nx)\) and \(\cos(nx)\)

We have by binomial expansion,

\(( cos \ x + isin \ x )^ n = \sum_{k=0}^{n} {n \choose k} (cos \ x )^k (sin \ x )^{n-k} \ i^{n-k}\)

We have,

\(i^{n-k} = (cos \ \frac{\pi}{2} + i sin \ \frac{\pi}{2} )^{n-k}\)

= \(cos \ (n-k) \frac{\pi}{2} + i sin \ (n-k) \frac{\pi}{2}\)

Therefore,

\(( cos \ x + isin \ x )^ n\)

= \(\sum_{k=0}^{n} {n \choose k} (cos \ x )^k (sin \ x )^{n-k} \ ( cos \ (n-k) \frac{\pi}{2} + i sin \ (n-k) \frac{\pi}{2} )\)

= \(cos \ nx + isin \ nx\)

Equating Real and Imaginary parts we get,

\(cos \ nx = \sum_{k=0}^{n} {n \choose k} (cos \ x )^k (sin \ x )^{n-k} \ cos \ (n-k) \frac{\pi}{2}\)

and,

\(sin \ nx = \sum_{k=0}^{n} {n \choose k} (cos \ x )^k (sin \ x )^{n-k} \ sin \ (n-k) \frac{\pi}{2}\)

Hence, we found very useful formulae for sine and cosine using De Moivre’s formula.

Finding \(n^{th}\) Root of a Complex Number

Let \(z\) be a complex number.

To find: \(z^(\frac{1}{n} ) ; n = 1, 2, 3 …\)

Write z in the polar form.

\(z = r ( cos \ x + isin \ x )\)

let \(z^{\frac{1}{n} } = j\)

 \(z  = j^n\)

Let \(j = \beta ( cos \ \alpha + isin \ \alpha )\)

Using De Moivre’s formula,

\({\beta}^n ( cos \ n \alpha + isin \ n \alpha )  = r ( cos \ x + isin \ x )\)

Comparing terms,

\({\beta }^n = r\)

\(\beta = r ^{\frac{1}{n} }\)

\(cos \ n \alpha = cos \ x\)

\(n \alpha = x + 2 \pi k ; k = 0, 1, 2 … , n-1\)

\(\alpha = \frac{x + 2 \pi k}{n}\)

Therefore,

\(z^{ \frac{1}{n} } = ( r ( cos \ x + isin \ x ) )^{ \frac{1}{n} }\)

 = \(r^{ \frac{1}{n} } ( cos \ \frac{x + 2 \pi k }{n}  + isin \ \frac{x + 2 \pi k }{n} ) ; k = 0, 1, 2, … , n-1\)

Examples

Question 1. Simplify \(( \frac{  1+  cos 2 \ x +  isin 2 \ x }{ 1+  cos 2 \ x –  isin 2 \ x } )^{30}\)

Solution. Let \(z =  cos \ 2x +  isin \ 2 x\).

Therefore,

\(( \frac{  1+  cos \  2x +  isin \  2 x }{ 1+  cos \ 2x –  isin \ 2x } )^{30}\)

\(( \frac{1 + z}{ 1+ \overline{z}} )^{30}\) = \(( \frac{  (1 + z)}{ 1 + \frac{1}{z}} )^{30}\)

= \(( \frac{ ( 1 + z)z}{ 1+ z})^{30}\) = \(( z )^{30}\) = \(( cos \ 2x +  isin \ 2 x )^{30}\)

= \(cos \ 60x + isin \ 60x\).

Question 2. Find All the cube roots of unity.

Solution. \(z = 1 = cos \ 0 + isin \ 0\)

Using the formula,

\(z^{ \frac{1}{n} } = ( r ( cos \ x + isin \ x ) )^{ \frac{1}{n} }\)

 = \(r^{ \frac{1}{n} } ( cos \ \frac{x + 2 \pi k }{n}  + isin \ \frac{x + 2 \pi k }{n} ) ; k = 1,2,… n-1\)

We have,

\(1^{ \frac{1}{3} } = (  cos \ 0 + isin \ 0  )^{ \frac{1}{3} }\)

 = \(( cos \ \frac{0 + 2 \pi k }{3}  + isin \ \frac{0 + 2 \pi k }{3} ) ; k = 0, 1, 2.\)

= \(( cos \ \frac{ 2 \pi k }{3}  + isin \ \frac{2 \pi k }{3} ) ; k = 0, 1, 2.\)

For \(k = 0\):

\(1^{ \frac{1}{3} } = 1\)

For \(k = 1\):

\(1^{ \frac{1}{3} }  = ( cos \ \frac{ 2 \pi  }{3}  + isin \ \frac{2 \pi  }{3} )\)

= \(( cos \ \pi –  \frac{  \pi }{3}  + isin \ \pi – \frac{\pi }{3} )\) = \(( –  cos \ \frac{ \pi }{3}  + isin \ \frac{\pi }{3} )\)

= \(- \frac{1}{2}  + i \frac{\sqrt{3}}{2}\)

For \(k = 2\):

\(1^{ \frac{1}{3} }  = ( cos \ \frac{ 4 \pi  }{3}  + isin \ \frac{4 \pi  }{3} )\)

= \(( cos \ \pi +  \frac{  \pi }{3}  + isin \ \pi + \frac{\pi }{3} )\) = \(( –  cos \ \frac{ \pi }{3}  –  isin \ \frac{\pi }{3} )\)

= \(- \frac{1}{2}  –  i \frac{\sqrt{3}}{2}\)

Therefore Cube roots of unity are: \(1\), \(- \frac{1}{2}  + i \frac{\sqrt{3}}{2}\) and \(- \frac{1}{2}  –  i \frac{\sqrt{3}}{2}\).

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