# Cube Root of Unity

Lets first generalize the concept of cube root of unit by nth root of Unity

## nth Roots of Unity

Let us take the equation
zn =1 , Here n is positive number
Mathematically this equation should be nth roots

### How to Find the nth roots

Now $1 = cos 0 + i sin 0$
So ,$z^n =cos 0 + i sin 0$
or
$z= (cos 0 + i sin 0)^{\frac {1}{n}}$

By De Moivre's theorem
$z= (cos \frac {2kπ}{n} + i sin \frac {2kπ}{n})$ and m=0,1,2,..n-1

Now It can be written in Euler form as
$z=e^{\frac {i2 \pi k}{n}}$
or
$z= \omega ^k$ where $\omega= e^{\frac {i2 \pi }{n}}$

Therefore, nth roots are
For $k=0,z^0=1$
$k=1,z^1=\omega$
$k=2,z^2=\omega ^ 2$
....
$k=n-1, z^n=\omega ^{n-1}$

Therefore, nth roots of unity are $1,\omega,\omega ^2,\omega ^3,....,\omega ^{n-1}$

### Properties of nth roots of Unity

• Sum of nth roots of unity is given by
$1+\omega+ \omega ^2 +\omega ^3+.....+ \omega ^{n-1}$
Now we can observe that it is geometric series having first term 1 and common ratio ω, So By using sum of n terms of a G.P

$1+\omega+ \omega ^2 +\omega ^3+.....+ \omega ^{n-1}$
$=\frac { 1 - \omega ^n}{ 1 - \omega}$

Now Since $\omega$ is nth root of unity, $\omega ^n = 1$
Therefore, $1+\omega;+ \omega ^2 +\omega ^3+.....+ \omega ^{n-1} = 0$
• Product of the nth roots of any complex number z is $(-1)^{n-1}$
• nth root of unity lies on the unit circle |z|=1 in the Argand plane and it divides the circle in n parts
• if $1, \alpha, \alpha ^2,\alpha ^3,......,\alpha ^{n-1}$ are the nth root of unity,then
$(1 - \alpha)(1 - \alpha ^2).......(1 - \alpha ^{n-1}) = n$ and,
$(1 + \alpha)(1 + \alpha ^2).......(1 + \alpha ^{n-1})$
= 0 if n is even and
= 1 if n is odd.
Now that we have studied the general concept,lets apply it to cube root of Unit

## Cube Root Of Unity

In the above generalization , if we put n=3, then
$\omega= e^{\frac {i2 \pi k}{3}}$ and The cube roots are $1,\omega, \omega ^2$
Now
$\omega$ value can be calculated as
$\omega =cos \frac {2 \pi}{3} + i sin \frac {2 \pi}{3}= \frac {-1+i \sqrt {3}}{2}$
So cube roots are 1, $\frac {-1+i \sqrt {3}}{2}$ and $\frac {-1-i \sqrt {3}}{2}$

We can observe that one root is real and other two are complex roots and they are conjugate to each other

### Properties of Cube roots of Unity

• $z^3 -1 =(z-1)(z-\omega)(z-\omega ^2)$
• $\omega$ and $\omega ^2$ are roots of the equation $z^2 +z + 1=0$
• Sum of the roots is $1+ \omega + \omega ^2 =0$
• Products of the roots
$1 \times \omega \times \omega ^2 =\omega ^3=1$
This is an important property
Lets see some more examples
$\omega ^4 =\omega ^3 \omega =\omega$
$\omega ^5=\omega ^3 \omega ^2=\omega ^2$
$\omega ^6=\omega ^3 \omega ^3=1$
So we can generalized this as
$\omega ^{3n} = 1$

$\omega ^{3n+1} = \omega ^{3n} \times \omega = \omega$
$\omega ^{3n+2}= \omega ^{3n} \times \omega ^2 = \omega ^2$
• $1 + \omega ^n +\omega ^{2n} =0$ if n is not a multiple of 3
$1 + \omega ^n +\omega ^{2n} =3$ if n is multiple of 3
• Cube roots of -1 are $-1, -\omega,\omega ^2$
• $z^3 +1 =(z+1)(z+\omega)(z+\omega ^2)$
• The following factorization should be remembered:
(a, b, c are Real Numbers and ω is the cube root of unity)
$a^3 - b^3= (a-b) (a-\omega b) (a-\omega ^2 b)$
$x^2 +x+1=(x-\omega)(x-\omega ^2)$
$a^3 - b^3= (a+b) (a+\omega b) (a+\omega ^2 b)$
$a^2 +ab+b^2 = (a-b \omega)(a-b \omega ^2)$
$a^3 + b^3 + c^3 - 3abc = (a+b+c) (a+\omega b+\omega ^2 c) (a+\omega ^2 b+\omega c)$

## Solved Examples

1) Find the value of expression
$1.(2-\omega)(2-\omega ^2) + 2.(3-\omega)(3-\omega ^2)+ .....(n-1)(3-\omega)(3-\omega ^2)$
Where $\omega$ is cube root of Unity

Solution
We know that
$z^3 -1 =(z-1)(z-\omega)(z-\omega ^2)$
So above expression can be written as
$\sum _{r=2}^{n}(r-1)(r-\omega )(r-\omega ^2)$
$=\sum _{r=2}^{n}(r^3 -1)$
$=\sum _{r=2}^{n}r^3 - \sum _{r=2}^{n}1$
$=( \frac {n(n+1)}{2})^2 -1 -(n-1) =( \frac {n(n+1)}{2})^2 -n$

2) (1+i)8 +(1-i)8
Solution
1+ i =√2( cos π/4 + isin π/4) and 1-i =√2( cos π/4 - isin π/4)

Therefore,
(1+i)8 +(1-i)8= 16( cos π/4 + isin π/4)8 + 16(cos π/4 - isin π/4)8
=16(cos 2π + sin 2π) + 16(cos 2π - sin 2π)
=32

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