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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 : 42 of 120

Marks: +1, -0

what is the value of

\rm \frac{1}{2}+Re\left(\frac{Z_1}{Z_2}\right)?

Solution:

Concept:

The problem uses the cube roots of unity,

\omega

and

\omega^2

, where

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

and

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

, with

\omega^3 = 1

and

1+\omega+\omega^2 = 0

Explanation:

Assume

Z_1 = \omega

and

Z_2 = \omega^2

(typical assignment).

Then

\frac{Z_1}{Z_2} = \frac{\omega}{\omega^2} = \omega^{-1} = \omega^2

because

\omega^3=1

The real part of

\omega^2

-\frac12

Therefore,

\frac12 + \operatorname{Re}\left(\frac{Z_1}{Z_2}\right) = \frac12 + \left(-\frac12\right) = 0

Answer:

Option B:

0

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