# Euler's Formula And De Moivre's Theorem

Euler’s formula:

Euler’s formula states that ‘For any real number $x$, $e^{ix}$ = $cos~x~+~i ~sin~x$.

Let z be a non zero complex number; we can write $z$ in the polar form as,

$z$ = $r(cos~θ~+~i~ sin~θ)$ = $r~e^{iθ}$, where $r$ is the modulus and $θ$ is argument of $z$.

• Multiplying a complex number $z$ with $e^{iα}$ gives, $ze^{iα}$ = $re^{iθ}~×~e^{iα}$ = $re^{i(α~+~θ)}$The resulting complex number $re^{i(α+θ)}$ will have the same modulus $r$ and argument $(α+θ)$.

De Moivre’s theorem:

It states that for any integer $n$,

$(cos ~θ~+~i ~sin~ θ)^n$ = $cos~ (nθ)~+~i ~sin~ (nθ)$

This can be easily proved using Euler’s formula as shown below.

We know that, $(cos~ θ~+~i ~sin ~θ)$ = $e^{iθ}$

$(cos~ θ~+~i ~sin~ θ)^n$ = $e^{i(nθ)}$

Therefore,

$e^{i(nθ)}$ = $cos ~(nθ)~+~i~ sin~ (nθ)$

$n^{th}$ roots of unity

If any complex number satisfies the equation $z^n$ = $1$, it is known as $n^{th}$ root of unity.

Fundamental theorem of algebra says that, an equation of degree $n$ will have $n$ roots. Therefore, there are $n$ values of $z$ which satisfies $z^n$ = $1$.

To find the values of $z$, we can write,

$1$ = $cos ~(2kπ)~ + ~i ~sin~ (2kπ)$, —(1) where k can be any integer.

We have,

$z^n$ = $1$

$z$ = $1^\frac{1}{n}$

From (1),

$z$ = $[cos~ (2kπ)~+~i~ sin~ (2kπ)]^{\frac{1}{n}}$

By De Moivre’s theorem,

$z$ = $[cos~ \left({2kπ}{n}\right)~+~i ~sin~ \left(\frac{2kπ}{n}\right)]$, where $k$ = $0, 1, 2, 3, …….., n-1$

For example; if $n$ = $3$, then $k$ = $0, 1, 2$

We know that, $z$ = $cos~ \left({2kπ}{n}\right)~+~i ~sin~ \left(\frac{2kπ}{n}\right)$ = $e^{\frac{2kπi}{n}}$

Let $ω$ = $cos~ \left(\frac{2π}{n}\right)~+i~ sin ~\left(\frac{2π}{n}\right)$ = $e^{\frac{2πi}{n}}$

$n^{th}$ roots of unity are found by,

When $k$ = $0$; $z$ = $1$

$k$ = $1$; $z$ = $ω$

$k$ = $2$; $z$ = $ω^2$

$k$ = $n$; $z$ = $ω^{n~-~1}$

Therefore, $n^{th}$ roots of unity are $1, ω, ω^2, ω^3,…….,ω^{n~-~1}$

• Sum of $n^{th}$ roots of unity is,$1~+~ω~+~ω^2~+~ω^3~+~⋯~+~ω^{n~-~1}$It is geometric series having first term 1 and common ratio $ω$.By using sum of $n$ terms of a G.P,$1~+~ω~+~ω^2~+~ω^3~+~⋯~+~ω^{n~-~1}$ = $\frac{1~-~ω^n}{1~-~ω}$Since $ω$ is $n^{th}$ root of unity, $ω^n$ = $1$Therefore, $1~+~ω~+~ω^2~+~ω^3~+~⋯~+~ω^{n~-~1}$ = $0$

Cube roots of unity:

We know that $n^{th}$ roots of unity are $1, ω, ω^2, ω^3,…….,ω^{n~-~1}$.

Therefore, cube roots of unity are $1, ω, ω^2$ where,

$ω$ = $cos ~\left(\frac{2π}{3}\right)~+~i~ sin~ \left(\frac{2π}{3}\right)$ = $\frac{-1~+~√3~ i}{2}$

$ω^2$ = $cos \left(\frac{4π}{3}\right)~+~i~ sin~ \left(\frac{4π}{3}\right)$ = $\frac{-1~-~√3~ i}{2}$

Sum of the cube roots of the unity,

$1~+~ω~+~ω^2$ = $0$

Product of cube roots of the unity,

$1~×~ω~×~ω^2$ = $ω^3$ = $1$

Example: $a$ and $b$ are the roots of the equation $x^2~+~x~+~1$ = $0$, Find the value of $a^{17}~+~b^{20}$

Roots of the equation are

$a$ = $\frac{-1~+~√{1~-~4}}{2}$ = $\frac{-1~+~√{3i}}{2}$

$b$ = $\frac{-1~-~√3~i}{2}$

Values of $a$ and $b$ are equal to $ω$ and $ω^2$ respectively.

$a^{17}~+~b^{20}$ = $ω^{17}~+~(ω^2)^{20}$ = $ω^{17}~+~ω^{40}$ = $ω^2~+~ω$

[Since $ω^{17}$ = $ω^{15}~×~ω^2$ and $ω^{40}$ = $ω^{39}~×~ω$]

[And, since $1~+~ω~+~ω^2$ = $0$]

Therefore,

$a^{17}~+~b^{20}$ = $-1$<

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#### Practise This Question

If |z|=1, then (1+z1+¯z)n+(1+¯z1+z)n is equal to