z

Mathematically this equation should be nth roots

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 $1,\omega,\omega ^2,\omega ^3,....,\omega ^{n-1}$

- 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.

$ \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

- $z^3 -1 =(z-1)(z-\omega)(z-\omega ^2)$
- $\omega$ 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)

$a^3 - b^3= (a-b) (a-\omega b) (a-\omega ^2 b)$

$x^2 +x+1=(x-\omega)(x-\omega ^2)$

$a^3 - b^3= (a+b) (a+\omega b) (a+\omega ^2 b)$

$a^2 +ab+b^2 = (a-b \omega)(a-b \omega ^2)$

$a^3 + b^3 + c^3 - 3abc = (a+b+c) (a+\omega b+\omega ^2 c) (a+\omega ^2 b+\omega c)$

$1.(2-\omega)(2-\omega ^2) + 2.(3-\omega)(3-\omega ^2)+ .....(n-1)(3-\omega)(3-\omega ^2)$

Where $\omega$ is cube root of Unity

We know that

$z^3 -1 =(z-1)(z-\omega)(z-\omega ^2)$

So above expression can be written as

$\sum _{r=2}^{n}(r-1)(r-\omega )(r-\omega ^2)$

$=\sum _{r=2}^{n}(r^3 -1)$

$=\sum _{r=2}^{n}r^3 - \sum _{r=2}^{n}1$

$=( \frac {n(n+1)}{2})^2 -1 -(n-1) =( \frac {n(n+1)}{2})^2 -n$

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

Therefore,

(1+i)

=16(cos 2π + sin 2π) + 16(cos 2π - sin 2π)

=32

$a^{1/3},a^{1/3} \omega, a^{1/3} \omega ^2$

**Notes**- What is complex numbers
- Algebra Of complex Numbers
- Conjugate of Complex Numbers
- Modulus of complex numbers
- Argand Plane
- Polar Representation of the complex number
- Rotation of Complex Number
- Identities for Complex Numbers
- Eulers formula and De moivre's theorem
- Cube Root of unity
- Complex roots of Quadratic equations

**NCERT Solutions & Worksheets**