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UPSC NDA 2 2019 Mathematics Solved Paper

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Question : 18 of 120

Marks: +1, -0

If

\alpha

\beta

are the roots of

x^{2} + x + 1 = 0

\sum\limits_{j=0}^{3}(\alpha^{j}+\beta^{j})

equal to?

8
6
4
2

Solution:

Concept:

The roots of

x^2 + x + 1 = 0

are the complex cube roots of unity:

\omega

and

\omega^2

, where

\omega^3 = 1

and

1 + \omega + \omega^2 = 0

.

Explanation:

Let

\alpha = \omega

and

\beta = \omega^2

.

We need

\displaystyle\sum_{j=0}^{3} (\alpha^j+\beta^j) = (\alpha^0+\beta^0)+(\alpha^1+\beta^1)+(\alpha^2+\beta^2)+(\alpha^3+\beta^3)

.

Compute each term:

•

j=0

:

\alpha^0 = 1

,

\beta^0 = 1

, so sum =

2

.

•

j=1

:

\alpha^1 + \beta^1 = \omega + \omega^2 = -1

(since

1+\omega+\omega^2=0

).

•

j=2

:

\alpha^2 = \omega^2

,

\beta^2 = (\omega^2)^2 = \omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega

. So

\omega^2 + \omega = -1

.

•

j=3

:

\alpha^3 = \omega^3 = 1

,

\beta^3 = (\omega^2)^3 = \omega^6 = (\omega^3)^2 = 1^2 = 1

, sum =

2

.

Now add:

2 + (-1) + (-1) + 2 = 2

.

Answer:

2

(Option D)

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