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,
\(1\) = \(cos ~(2kÏ€)~ + ~i ~sin~ (2kÏ€)\), —(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 \(1~+~ω~+~ω^2\) = \(0\)]Therefore,
\(a^{17}~+~b^{20}\) = \(-1\)
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