Euler’s formula
Euler’s formula states that ‘For any real number \(\begin{array}{l}x\end{array} \)
, \(\begin{array}{l}e^{ix}\end{array} \)
= \(\begin{array}{l}cos~x~+~i ~sin~x\end{array} \)
.
Let z be a non zero complex number; we can write \(\begin{array}{l}z\end{array} \)
in the polar form as,
\(\begin{array}{l}z\end{array} \)
= \(\begin{array}{l}r(cos~θ~+~i~ sin~θ)\end{array} \)
= \(\begin{array}{l}r~e^{iθ}\end{array} \)
, where \(\begin{array}{l}r\end{array} \)
is the modulus and \(\begin{array}{l}θ\end{array} \)
is argument of \(\begin{array}{l}z\end{array} \)
.
Multiplying a complex number \(\begin{array}{l}z\end{array} \)
with \(\begin{array}{l}e^{iα}\end{array} \)
gives, \(\begin{array}{l}ze^{iα}\end{array} \)
= \(\begin{array}{l}re^{iθ}~×~e^{iα}\end{array} \)
= \(\begin{array}{l}re^{i(α~+~θ)}\end{array} \)
The resulting complex number \(\begin{array}{l}re^{i(α+θ)}\end{array} \)
will have the same modulus \(\begin{array}{l}r\end{array} \)
and argument \(\begin{array}{l}(α+θ)\end{array} \)
.
De Moivre’s theorem
It states that for any integer \(\begin{array}{l}n\end{array} \)
,
\(\begin{array}{l}(cos ~θ~+~i ~sin~ θ)^n\end{array} \)
= \(\begin{array}{l}cos~ (nθ)~+~i ~sin~ (nθ)\end{array} \)
This can be easily proved using Euler’s formula as shown below.
We know that, \(\begin{array}{l}(cos~ θ~+~i ~sin ~θ)\end{array} \)
= \(\begin{array}{l}e^{iθ}\end{array} \)
\(\begin{array}{l}(cos~ θ~+~i ~sin~ θ)^n\end{array} \)
= \(\begin{array}{l}e^{i(nθ)}\end{array} \)
Therefore,
\(\begin{array}{l}e^{i(nθ)}\end{array} \)
= \(\begin{array}{l}cos ~(nθ)~+~i~ sin~ (nθ)\end{array} \)
\(\begin{array}{l}n^{th}\end{array} \)
roots of unity
If any complex number satisfies the equation \(\begin{array}{l}z^n\end{array} \)
= \(\begin{array}{l}1\end{array} \)
, it is known as \(\begin{array}{l}n^{th}\end{array} \)
root of unity.
Fundamental theorem of algebra says that, an equation of degree \(\begin{array}{l}n\end{array} \)
will have \(\begin{array}{l}n\end{array} \)
roots. Therefore, there are \(\begin{array}{l}n\end{array} \)
values of \(\begin{array}{l}z\end{array} \)
which satisfies \(\begin{array}{l}z^n\end{array} \)
= \(\begin{array}{l}1\end{array} \)
.
To find the values of \(\begin{array}{l}z\end{array} \)
, we can write,
\(\begin{array}{l}1\end{array} \)
= \(\begin{array}{l}cos ~(2kπ)~ + ~i ~sin~ (2kπ)\end{array} \)
, —(1) where k can be any integer.
We have,
\(\begin{array}{l}z^n\end{array} \)
= \(\begin{array}{l}1\end{array} \)
\(\begin{array}{l}z\end{array} \)
= \(\begin{array}{l}1^\frac{1}{n}\end{array} \)
From (1),
\(\begin{array}{l}z\end{array} \)
= \(\begin{array}{l}[cos~ (2kπ)~+~i~ sin~ (2kπ)]^{\frac{1}{n}}\end{array} \)
By De Moivre’s theorem,
\(\begin{array}{l}z\end{array} \)
= \(\begin{array}{l}[cos~ \left(\frac{2kπ}{n}\right)~+~i ~sin~ \left(\frac{2kπ}{n}\right)]\end{array} \)
, where \(\begin{array}{l}k\end{array} \)
= \(\begin{array}{l}0, 1, 2, 3, …….., n-1\end{array} \)
For example; if \(\begin{array}{l}n\end{array} \)
= \(\begin{array}{l}3\end{array} \)
, then \(\begin{array}{l}k\end{array} \)
= \(\begin{array}{l}0, 1, 2\end{array} \)
We know that, \(\begin{array}{l}z\end{array} \)
= \(\begin{array}{l}cos~ \left(\frac{2kπ}{n}\right)~+~i ~sin~ \left(\frac{2kπ}{n}\right)\end{array} \)
= \(\begin{array}{l}e^{\frac{2kπi}{n}}\end{array} \)
Let \(\begin{array}{l}ω\end{array} \)
= \(\begin{array}{l}cos~ \left(\frac{2Ï€}{n}\right)~+i~ sin ~\left(\frac{2Ï€}{n}\right)\end{array} \)
= \(\begin{array}{l}e^{\frac{2Ï€i}{n}}\end{array} \)
\(\begin{array}{l}n^{th}\end{array} \)
roots of unity are found by,
When \(\begin{array}{l}k\end{array} \)
= \(\begin{array}{l}0\end{array} \)
; \(\begin{array}{l}z\end{array} \)
= \(\begin{array}{l}1\end{array} \)
\(\begin{array}{l}k\end{array} \)
= \(\begin{array}{l}1\end{array} \)
; \(\begin{array}{l}z\end{array} \)
= \(\begin{array}{l}ω\end{array} \)
\(\begin{array}{l}k\end{array} \)
= \(\begin{array}{l}2\end{array} \)
; \(\begin{array}{l}z\end{array} \)
= \(\begin{array}{l}ω^2\end{array} \)
\(\begin{array}{l}k\end{array} \)
= \(\begin{array}{l}n\end{array} \)
; \(\begin{array}{l}z\end{array} \)
= \(\begin{array}{l}ω^{n~-~1}\end{array} \)
Therefore, \(\begin{array}{l}n^{th}\end{array} \)
roots of unity are \(\begin{array}{l}1, ω, ω^2, ω^3,…….,ω^{n~-~1}\end{array} \)
- Sum of
\(\begin{array}{l}n^{th}\end{array} \)
roots of unity is,\(\begin{array}{l}1~+~ω~+~ω^2~+~ω^3~+~⋯~+~ω^{n~-~1}\end{array} \)
It is geometric series having first term 1 and common ratio \(\begin{array}{l}ω\end{array} \)
.By using sum of \(\begin{array}{l}n\end{array} \)
terms of a G.P,\(\begin{array}{l}1~+~ω~+~ω^2~+~ω^3~+~⋯~+~ω^{n~-~1}\end{array} \)
= \(\begin{array}{l}\frac{1~-~ω^n}{1~-~ω}\end{array} \)
Since \(\begin{array}{l}ω\end{array} \)
is \(\begin{array}{l}n^{th}\end{array} \)
root of unity, \(\begin{array}{l}ω^n\end{array} \)
= \(\begin{array}{l}1\end{array} \)
Therefore, \(\begin{array}{l}1~+~ω~+~ω^2~+~ω^3~+~⋯~+~ω^{n~-~1}\end{array} \)
= \(\begin{array}{l}0\end{array} \)
Cube roots of unity:
We know that \(\begin{array}{l}n^{th}\end{array} \)
roots of unity are \(\begin{array}{l}1, ω, ω^2, ω^3,…….,ω^{n~-~1}\end{array} \)
.
Therefore, cube roots of unity are \(\begin{array}{l}1, ω, ω^2\end{array} \)
where,
\(\begin{array}{l}ω\end{array} \)
= \(\begin{array}{l}cos ~\left(\frac{2Ï€}{3}\right)~+~i~ sin~ \left(\frac{2Ï€}{3}\right)\end{array} \)
= \(\begin{array}{l}\frac{-1~+~√3~ i}{2}\end{array} \)
\(\begin{array}{l}ω^2\end{array} \)
= \(\begin{array}{l}cos \left(\frac{4Ï€}{3}\right)~+~i~ sin~ \left(\frac{4Ï€}{3}\right)\end{array} \)
= \(\begin{array}{l}\frac{-1~-~√3~ i}{2}\end{array} \)
Sum of the cube roots of the unity,
\(\begin{array}{l}1~+~ω~+~ω^2\end{array} \)
= \(\begin{array}{l}0\end{array} \)
Product of cube roots of the unity,
\(\begin{array}{l}1~×~ω~×~ω^2\end{array} \)
= \(\begin{array}{l}ω^3\end{array} \)
= \(\begin{array}{l}1\end{array} \)
Example: \(\begin{array}{l}a\end{array} \)
and \(\begin{array}{l}b\end{array} \)
are the roots of the equation \(\begin{array}{l}x^2~+~x~+~1\end{array} \)
= \(\begin{array}{l}0\end{array} \)
, Find the value of \(\begin{array}{l}a^{17}~+~b^{20}\end{array} \)
Roots of the equation are
\(\begin{array}{l}a\end{array} \)
= \(\begin{array}{l}\frac{-1~+~√({1~-~4})}{2}\end{array} \)
= \(\begin{array}{l}\frac{-1~+~√{3i}}{2}\end{array} \)
\(\begin{array}{l}b\end{array} \)
= \(\begin{array}{l}\frac{-1~-~√3~i}{2}\end{array} \)
Values of \(\begin{array}{l}a\end{array} \)
and \(\begin{array}{l}b\end{array} \)
are equal to \(\begin{array}{l}ω\end{array} \)
and \(\begin{array}{l}ω^2\end{array} \)
respectively.
\(\begin{array}{l}a^{17}~+~b^{20}\end{array} \)
= \(\begin{array}{l}ω^{17}~+~(ω^2)^{20}\end{array} \)
= \(\begin{array}{l}ω^{17}~+~ω^{40}\end{array} \)
= \(\begin{array}{l}ω^2~+~ω\end{array} \)
[Since \(\begin{array}{l}ω^{17}\end{array} \)
= \(\begin{array}{l}ω^{15}~×~ω^2\end{array} \)
and \(\begin{array}{l}ω^{40}\end{array} \)
= \(\begin{array}{l}ω^{39}~×~ω\end{array} \)
]
[And, since \(\begin{array}{l}1~+~ω~+~ω^2\end{array} \)
= \(\begin{array}{l}0\end{array} \)
]
Therefore,
\(\begin{array}{l}a^{17}~+~b^{20}\end{array} \)
= \(\begin{array}{l}-1\end{array} \)
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