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UPSC NDA I 2019 Mathematics Solved Paper

Question Numbers: 29-30

Direction: Consider the following for the next 02 (two) items:

A complex number is given by z.

z = \frac{{1 + 2i}}{{1 - {{\left( {1 - i} \right)}^2}}}

Question : 29 of 120

Marks: +1, -0

What is the modulus of z?

Solution:

Concept:

Modulus of a complex number

z = a + ib

|z| = \sqrt{a^2 + b^2}

Explanation:

Given:

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

First simplify the denominator:

(1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i

So denominator becomes

1 - (-2i) = 1 + 2i

Thus

z = \frac{1 + 2i}{1 + 2i} = 1

Now

z = 1 = 1 + 0 \cdot i

, so

a = 1

b = 0

Modulus

|z| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1

Answer:

C. 1

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