Question

Plzz answer it quick by solving steps.

Answer 1

Answer:

Answer is 2.

Option => B

Step-by-step explanation:

This question is all about complex numbers,

Complex numbers:

A number that can be showed/expressed in type/form of  + bi, where a & b are real numbers.

We know that,

Cute root of unity:

Where,

x³ = 1 and, x = 1,$\sf \dfrac{-1+i\sqrt{3}}{2} = \omega,\dfrac{-i,-1\sqrt{3}}{2} = \omega^2$

$(Omega\;is\;the\;root\;of\;x^3 = 1)\\\ So,that\;means\; \omega^3 =1$

Here,

Similarly:

$=1 + \omega +\omega^3 = 0$

According to your question:

We have to square this equation

$\sf \bigg(\dfrac{-1+i\sqrt{3}}{2}\bigg)^3n$

With omega,

So we will get,

$\omega^3^n + (\omega^2)^3^n$

We can write it as,

$(\omega^2)^3^n + (\omega^3})^2^n$

After substituting these values,

We get,

$1^n +1^2^n = 1+1 = 2\\\\\ => 2\quad(Answer)$

$\rule{300}{3.5}$

Answer 2

Answer:-

2

Given :-

$(\dfrac{-1 + i\sqrt{3}}{2}) ^{3n} + (\dfrac{-1-i\sqrt{3}}{2}) ^{3n}$

To find :-

It's value.

Solution:-

We know that every cubic polynomial has three roots and 1 have also three roots.

That is, 1 $\omega$ and $\omega^2$

The relationship between them is :-

$\huge \boxed {\omega + \omega ^2 + 1 = 0}$

Where,

$\omega = \dfrac{-1+i\sqrt{3}}{2}$

$\omega ^2 = \dfrac{-1 -i\sqrt{3}}{2}$

Also,

$\huge \boxed{\omega ^3 = 1}$

As per question,

→$(\omega)^{3n}+ (\omega^2 )^{3n}$

→$(\omega ^2)^{3n} +(\omega^3 )^{2n}$

→$(1)^n + (1^2)^n$

→$2 \times 1^n$

→$2$

The power of 1 is always 1.

hence,

The correct answer will be 2.