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UPSC NDA II 2018 Mathematics Solved Paper

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Question : 15 of 120

Marks: +1, -0

What is the value of

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

i=\sqrt{-1}?

3
2
1
0

Solution:

Concept:

The given expressions are the complex cube roots of unity, denoted as

\omega

and

\omega^2

.

Explanation:

Let

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

and

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

.

These satisfy

\omega^3 = 1

.

Thus,

\left( \frac{-1 + i\sqrt{3}}{2} \right)^{3n} = \omega^{3n} = (\omega^3)^n = 1^n = 1

.

Similarly,

\left( \frac{-1 - i\sqrt{3}}{2} \right)^{3n} = (\omega^2)^{3n} = \omega^{6n} = (\omega^3)^{2n} = 1^{2n} = 1

.

Adding them gives

1 + 1 = 2

.

Answer:

2

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