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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 33-35

Consider the following for the next items that follow:

Consider equation-I : z3 + 2z2 + 2z + 1 = 0 and equation-II : z1985 + z100 + 1 = 0.

Question : 33 of 120

Marks: +1, -0

What are the roots of equation-I ?

Solution:

Concept:

The cube roots of unity are

1

\omega

, and

\omega^2

with

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

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

, and properties

1 + \omega + \omega^2 = 0

\omega^3 = 1

Explanation:

The given cubic equation is

z^3 + 2z^2 + 2z + 1 = 0

Factorise the polynomial:

(z + 1)(z^2 + z + 1) = 0

Set each factor to zero:

z + 1 = 0

gives

z = -1

The quadratic

z^2 + z + 1 = 0

has roots

\omega

and

\omega^2

(the non‑real cube roots of unity).

Thus the three roots of the equation are

-1

\omega

, and

\omega^2

Answer:

The roots are

-1

\omega

, and

\omega^2

, which corresponds to option B.

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