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# Euler's Formula And De Moivre's Theorem

## Eulerâ€™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,

$$\begin{array}{l}z\end{array}$$
=
$$\begin{array}{l}r(cos~Î¸~+~i~ sin~Î¸)\end{array}$$
=
$$\begin{array}{l}r~e^{iÎ¸}\end{array}$$
, where
$$\begin{array}{l}r\end{array}$$
is the modulus and
$$\begin{array}{l}Î¸\end{array}$$
is argument of
$$\begin{array}{l}z\end{array}$$
.

Multiplying a complex number

$$\begin{array}{l}z\end{array}$$
with
$$\begin{array}{l}e^{iÎ±}\end{array}$$
gives,
$$\begin{array}{l}ze^{iÎ±}\end{array}$$
=
$$\begin{array}{l}re^{iÎ¸}~Ã—~e^{iÎ±}\end{array}$$
=
$$\begin{array}{l}re^{i(Î±~+~Î¸)}\end{array}$$
The resulting complex number
$$\begin{array}{l}re^{i(Î±+Î¸)}\end{array}$$
will have the same modulus
$$\begin{array}{l}r\end{array}$$
and argument
$$\begin{array}{l}(Î±+Î¸)\end{array}$$
.

## De Moivreâ€™s theorem

It states that for any integer

$$\begin{array}{l}n\end{array}$$
,

$$\begin{array}{l}(cos ~Î¸~+~i ~sin~ Î¸)^n\end{array}$$
=
$$\begin{array}{l}cos~ (nÎ¸)~+~i ~sin~ (nÎ¸)\end{array}$$

This can be easily proved using Eulerâ€™s formula as shown below.

We know that,

$$\begin{array}{l}(cos~ Î¸~+~i ~sin ~Î¸)\end{array}$$
=
$$\begin{array}{l}e^{iÎ¸}\end{array}$$

$$\begin{array}{l}(cos~ Î¸~+~i ~sin~ Î¸)^n\end{array}$$
=
$$\begin{array}{l}e^{i(nÎ¸)}\end{array}$$

Therefore,

$$\begin{array}{l}e^{i(nÎ¸)}\end{array}$$
=
$$\begin{array}{l}cos ~(nÎ¸)~+~i~ sin~ (nÎ¸)\end{array}$$

$$\begin{array}{l}n^{th}\end{array}$$
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,

$$\begin{array}{l}1\end{array}$$
=
$$\begin{array}{l}cos ~(2kÏ€)~ + ~i ~sin~ (2kÏ€)\end{array}$$
, —(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),

$$\begin{array}{l}z\end{array}$$
=
$$\begin{array}{l}[cos~ (2kÏ€)~+~i~ sin~ (2kÏ€)]^{\frac{1}{n}}\end{array}$$

By De Moivreâ€™s theorem,

$$\begin{array}{l}z\end{array}$$
=
$$\begin{array}{l}[cos~ \left(\frac{2kÏ€}{n}\right)~+~i ~sin~ \left(\frac{2kÏ€}{n}\right)]\end{array}$$
, where
$$\begin{array}{l}k\end{array}$$
=
$$\begin{array}{l}0, 1, 2, 3, â€¦â€¦.., n-1\end{array}$$

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,

$$\begin{array}{l}z\end{array}$$
=
$$\begin{array}{l}cos~ \left(\frac{2kÏ€}{n}\right)~+~i ~sin~ \left(\frac{2kÏ€}{n}\right)\end{array}$$
=
$$\begin{array}{l}e^{\frac{2kÏ€i}{n}}\end{array}$$

Let

$$\begin{array}{l}Ï‰\end{array}$$
=
$$\begin{array}{l}cos~ \left(\frac{2Ï€}{n}\right)~+i~ sin ~\left(\frac{2Ï€}{n}\right)\end{array}$$
=
$$\begin{array}{l}e^{\frac{2Ï€i}{n}}\end{array}$$

$$\begin{array}{l}n^{th}\end{array}$$
roots of unity are found by,

When

$$\begin{array}{l}k\end{array}$$
=
$$\begin{array}{l}0\end{array}$$
;
$$\begin{array}{l}z\end{array}$$
=
$$\begin{array}{l}1\end{array}$$

$$\begin{array}{l}k\end{array}$$
=
$$\begin{array}{l}1\end{array}$$
;
$$\begin{array}{l}z\end{array}$$
=
$$\begin{array}{l}Ï‰\end{array}$$

$$\begin{array}{l}k\end{array}$$
=
$$\begin{array}{l}2\end{array}$$
;
$$\begin{array}{l}z\end{array}$$
=
$$\begin{array}{l}Ï‰^2\end{array}$$

$$\begin{array}{l}k\end{array}$$
=
$$\begin{array}{l}n\end{array}$$
;
$$\begin{array}{l}z\end{array}$$
=
$$\begin{array}{l}Ï‰^{n~-~1}\end{array}$$

Therefore,

$$\begin{array}{l}n^{th}\end{array}$$
roots of unity are
$$\begin{array}{l}1, Ï‰, Ï‰^2, Ï‰^3,â€¦â€¦.,Ï‰^{n~-~1}\end{array}$$

• Sum of
$$\begin{array}{l}n^{th}\end{array}$$
roots of unity is,
$$\begin{array}{l}1~+~Ï‰~+~Ï‰^2~+~Ï‰^3~+~â‹¯~+~Ï‰^{n~-~1}\end{array}$$
It is geometric series having first term 1 and common ratio
$$\begin{array}{l}Ï‰\end{array}$$
.By using sum of
$$\begin{array}{l}n\end{array}$$
terms of a G.P,
$$\begin{array}{l}1~+~Ï‰~+~Ï‰^2~+~Ï‰^3~+~â‹¯~+~Ï‰^{n~-~1}\end{array}$$
=
$$\begin{array}{l}\frac{1~-~Ï‰^n}{1~-~Ï‰}\end{array}$$
Since
$$\begin{array}{l}Ï‰\end{array}$$
is
$$\begin{array}{l}n^{th}\end{array}$$
root of unity,
$$\begin{array}{l}Ï‰^n\end{array}$$
=
$$\begin{array}{l}1\end{array}$$
Therefore,
$$\begin{array}{l}1~+~Ï‰~+~Ï‰^2~+~Ï‰^3~+~â‹¯~+~Ï‰^{n~-~1}\end{array}$$
=
$$\begin{array}{l}0\end{array}$$

Cube roots of unity:

We know that

$$\begin{array}{l}n^{th}\end{array}$$
roots of unity are
$$\begin{array}{l}1, Ï‰, Ï‰^2, Ï‰^3,â€¦â€¦.,Ï‰^{n~-~1}\end{array}$$
.

Therefore, cube roots of unity are

$$\begin{array}{l}1, Ï‰, Ï‰^2\end{array}$$
where,

$$\begin{array}{l}Ï‰\end{array}$$
=
$$\begin{array}{l}cos ~\left(\frac{2Ï€}{3}\right)~+~i~ sin~ \left(\frac{2Ï€}{3}\right)\end{array}$$
=
$$\begin{array}{l}\frac{-1~+~âˆš3~ i}{2}\end{array}$$

$$\begin{array}{l}Ï‰^2\end{array}$$
=
$$\begin{array}{l}cos \left(\frac{4Ï€}{3}\right)~+~i~ sin~ \left(\frac{4Ï€}{3}\right)\end{array}$$
=
$$\begin{array}{l}\frac{-1~-~âˆš3~ i}{2}\end{array}$$

Sum of the cube roots of the unity,

$$\begin{array}{l}1~+~Ï‰~+~Ï‰^2\end{array}$$
=
$$\begin{array}{l}0\end{array}$$

Product of cube roots of the unity,

$$\begin{array}{l}1~Ã—~Ï‰~Ã—~Ï‰^2\end{array}$$
=
$$\begin{array}{l}Ï‰^3\end{array}$$
=
$$\begin{array}{l}1\end{array}$$

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

$$\begin{array}{l}a\end{array}$$
=
$$\begin{array}{l}\frac{-1~+~âˆš({1~-~4})}{2}\end{array}$$
=
$$\begin{array}{l}\frac{-1~+~âˆš{3i}}{2}\end{array}$$

$$\begin{array}{l}b\end{array}$$
=
$$\begin{array}{l}\frac{-1~-~âˆš3~i}{2}\end{array}$$

Values of

$$\begin{array}{l}a\end{array}$$
and
$$\begin{array}{l}b\end{array}$$
are equal to
$$\begin{array}{l}Ï‰\end{array}$$
and
$$\begin{array}{l}Ï‰^2\end{array}$$
respectively.

$$\begin{array}{l}a^{17}~+~b^{20}\end{array}$$
=
$$\begin{array}{l}Ï‰^{17}~+~(Ï‰^2)^{20}\end{array}$$
=
$$\begin{array}{l}Ï‰^{17}~+~Ï‰^{40}\end{array}$$
=
$$\begin{array}{l}Ï‰^2~+~Ï‰\end{array}$$

[Since

$$\begin{array}{l}Ï‰^{17}\end{array}$$
=
$$\begin{array}{l}Ï‰^{15}~Ã—~Ï‰^2\end{array}$$
and
$$\begin{array}{l}Ï‰^{40}\end{array}$$
=
$$\begin{array}{l}Ï‰^{39}~Ã—~Ï‰\end{array}$$
]

[And, since

$$\begin{array}{l}1~+~Ï‰~+~Ï‰^2\end{array}$$
=
$$\begin{array}{l}0\end{array}$$
]

Therefore,

$$\begin{array}{l}a^{17}~+~b^{20}\end{array}$$
=
$$\begin{array}{l}-1\end{array}$$

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