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UPSC NDA II 2024 Mathematics Paper

Question Numbers: 41-42

Direction : Consider the following for the items that follow :

Let

Z_{1}

and

Z_{2}

be any two complex numbers such that

rm Z_1^2+Z_2^2+Z_1Z_2=0

Question : 41 of 120

Marks: +1, -0

what is the value of

\rm \left | \frac{Z_1}{Z_2}\right |

Solution:

Concept:

Use cube roots of unity to determine the magnitude of the given ratio.

Explanation:

The given equation is

Z_1^2 + Z_2^2 + Z_1 Z_2 = 0

Divide both sides by

Z_2^2

(assuming

Z_2 \neq 0

\left(\frac{Z_1}{Z_2}\right)^2 + \left(\frac{Z_1}{Z_2}\right) + 1 = 0

Let

r = \frac{Z_1}{Z_2}

. Then

r^2 + r + 1 = 0

The roots of this quadratic are the complex cube roots of unity:

r = \omega

r = \omega^2

For any cube root of unity,

|\omega| = 1

and

|\omega^2| = 1

Therefore,

\left| \frac{Z_1}{Z_2} \right| = |r| = 1

Answer:

1

(Option A)

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