Question

1 - omega +omega square

Answer 1

Step-by-step explanation:

The values of \omegaω and \omega^2ω

2

are - (1 + \omegaω ) or \omega^2=\dfrac{-1-i\sqrt{3}}{2}ω

2

=

2

−1−i

3

and \omega^3=1ω

3

=1 .

Explanation:

We have,

\omegaω and \omega^2ω

2

To find, the value of \omegaω and \omega^2ω

2

= ?

We know that,

1, \omegaω and \omega^2ω

2

cube root of unity.

∴ 1 + \omegaω + \omega^2ω

2

= 0

⇒ \omega^2ω

2

= - (1 + \omegaω )

\omega=\dfrac{-1+i\sqrt{3}}{2}ω=

2

−1+i

3

and \omega^2=\dfrac{-1-i\sqrt{3}}{2}ω

2

=

2

−1−i

3

\omega^3=1ω

3

=1

∴ \omega^2ω

2

= - (1 + \omegaω ) or \omega^2=\dfrac{-1-i\sqrt{3}}{2}ω

2

=

2

−1−i

3

and \omega^3=1ω

3

=1

Thus, the values of \omegaω and \omega^2ω

2

are - (1 + \omegaω ) or \omega^2=\dfrac{-1-i\sqrt{3}}{2}ω

2

=

2

−1−i

3

and \omega^3=1ω

3

=1 .