Test Index

UPSC NDA 2 2016 Mathematics Solved Paper

© examsnet.com

Question : 5 of 120

Marks: +1, -0

What is

\omega^{100} + \omega^{200} + \omega^{300}

equal to, where

\omega

is the cube root of unity?

1
3ω
$3\omega^{2}$
0

Solution:

Concept:

The cube roots of unity are

1

,

\omega

, and

\omega^2

.

Key properties:

\omega^3 = 1

and

1 + \omega + \omega^2 = 0

.

Explanation:

We need to simplify

\omega^{100} + \omega^{200} + \omega^{300}

.

Write each power in terms of

\omega^3

.

\omega^{100} = \omega^{99} \cdot \omega = (\omega^3)^{33} \cdot \omega = 1 \cdot \omega = \omega

.

\omega^{200} = \omega^{198} \cdot \omega^2 = (\omega^3)^{66} \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2

.

\omega^{300} = (\omega^3)^{100} = 1

.

Now sum them:

\omega^{100} + \omega^{200} + \omega^{300} = \omega + \omega^2 + 1

.

Using the property

1 + \omega + \omega^2 = 0

, the sum equals

0

.

Answer:

The value is

0

, which corresponds to option D.

© examsnet.com

Go to Question:

Prev Question

Next Question