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Euler's Formula And De Moivre's Theorem

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
\(\begin{array}{l}(α+θ)\end{array} \)
.

Euler’s Formula And De Moivre’s Theorem

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

\(\begin{array}{l}ω\end{array} \)
=
\(\begin{array}{l}cos~ \left(\frac{2π}{n}\right)~+i~ sin ~\left(\frac{2π}{n}\right)\end{array} \)
=
\(\begin{array}{l}e^{\frac{2πi}{n}}\end{array} \)

\(\begin{array}{l}n^{th}\end{array} \)
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,

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

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

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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