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 \((α+θ)\).

Euler’s Formula And De Moivre’s Theorem

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, \((cos~ θ~+~i ~sin ~θ)\) = \(e^{iθ}\)

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


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

\(n^{th}\) 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,

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

For example; if \(n\) = \(3\), then \(k\) = \(0, 1, 2\)

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

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 \(1~+~ω~+~ω^2\) = \(0\)]


\(a^{17}~+~b^{20}\) = \(-1\)

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