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UPSC NDA 2 2022 Mathematics Solved Paper

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Question Numbers: 16-18

Let z =

\frac{1+\text{i}\sin \theta}{1-\text{i}\sin \theta}

where i =

\sqrt{-1}

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Question : 18 of 120

Marks: +1, -0

What is angle θ such that z is purely imaginary ?

where n is an integer

$\frac{n\pi}{2}$
$\frac{(2n+1)\pi}{2}$
$n\pi$
$2n\pi$

Solution:

Concept:

If a complex number

z = x + iy

is purely imaginary, its real part

x

is zero.

Explanation:

Given

z = \frac{1 + i \sin \theta}{1 - i \sin \theta}

.

Multiply numerator and denominator by

(1 + i \sin \theta)

:

z = \frac{(1 + i \sin \theta)^2}{1^2 - (i \sin \theta)^2} = \frac{1 + 2i \sin \theta - \sin^2 \theta}{1 + \sin^2 \theta}

.

Separate real and imaginary parts:

z = \frac{1 - \sin^2 \theta}{1 + \sin^2 \theta} + i \frac{2 \sin \theta}{1 + \sin^2 \theta} = \frac{\cos^2 \theta}{1 + \sin^2 \theta} + i \frac{2 \sin \theta}{1 + \sin^2 \theta}

.

For

z

to be purely imaginary, real part

= 0

:

\frac{\cos^2 \theta}{1 + \sin^2 \theta} = 0 \Rightarrow \cos^2 \theta = 0 \Rightarrow \cos \theta = 0

.

Hence

\theta = \frac{(2n+1)\pi}{2}

, where

n

is an integer.

Answer:

Option B:

\frac{(2n+1)\pi}{2}

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