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UPSC NDA II 2024 Mathematics Paper

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Question : 4 of 120

Marks: +1, -0

If

\omega \neq 1

is a cube root of unity, then what is

(1 + \omega - \omega^{2})^{100} + (1 - \omega + \omega^{2})^{100}

equal to?

$2^{100} \omega^{2}$
$2^{100} \omega$
$2^{100}$
$-2^{100}$

Solution:

Concept:

Use the properties of cube roots of unity:

1 + \omega + \omega^{2} = 0

and

\omega^{3} = 1

.

Explanation:

Given

1 + \omega + \omega^{2} = 0

, we have

1 + \omega = -\omega^{2}

and

1 + \omega^{2} = -\omega

.

Then

(1 + \omega - \omega^{2}) = (-\omega^{2}) - \omega^{2} = -2\omega^{2}

.

Similarly,

(1 - \omega + \omega^{2}) = (-\omega) - \omega = -2\omega

.

Now compute the sum:

(-2\omega^{2})^{100} + (-2\omega)^{100} = 2^{100}(\omega^{200} + \omega^{100})

.

Since

\omega^{3}=1

, we simplify:

\omega^{200} = \omega^{3 \times 66 + 2} = \omega^{2}

and

\omega^{100} = \omega^{3 \times 33 + 1} = \omega

.

Thus

\omega^{200} + \omega^{100} = \omega^{2} + \omega = -1

(because

1 + \omega + \omega^{2} = 0

).

Therefore the sum equals

2^{100} \times (-1) = -2^{100}

.

Answer:

-2^{100}

(Option D).

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