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UPSC NDA 2 2022 Mathematics Solved Paper

Question Numbers: 16-18

Let z =

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

where i =

\sqrt{-1}

Question : 16 of 120

Marks: +1, -0

What is the modulus of z?

Solution:

Concept:

For a complex number

z = x + \mathrm{i}y

, its modulus is given by

|z| = \sqrt{x^2 + y^2}

Explanation:

We are given

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

Multiply numerator and denominator by the conjugate

1 + \mathrm{i}\sin\theta

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

Expand the numerator:

(1 + \mathrm{i}\sin\theta)^2 = 1 - \sin^2\theta + 2\mathrm{i}\sin\theta = \cos^2\theta + 2\mathrm{i}\sin\theta

Thus

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

Now compute the modulus:

|z| = \sqrt{\left(\frac{\cos^2\theta}{1+\sin^2\theta}\right)^2 + \left(\frac{2\sin\theta}{1+\sin^2\theta}\right)^2}

= \frac{1}{1+\sin^2\theta}\sqrt{\cos^4\theta + 4\sin^2\theta}

Since

\cos^2\theta = 1 - \sin^2\theta

, we have

\cos^4\theta = (1 - \sin^2\theta)^2 = 1 - 2\sin^2\theta + \sin^4\theta

Then inside the square root:

(1 - 2\sin^2\theta + \sin^4\theta) + 4\sin^2\theta = 1 + 2\sin^2\theta + \sin^4\theta = (1+\sin^2\theta)^2

Therefore

|z| = \frac{1}{1+\sin^2\theta} \cdot (1+\sin^2\theta) = 1

Answer:

Option A:

1

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