Math, asked by nikhildhdwal09, 1 year ago

Plzz answer it quick by solving steps.

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Answers

Answered by Blaezii
18

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}

Answered by Anonymous
17

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.

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