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UPSC NDA 2 2017 Mathematics Solved Paper

Question : 21 of 120

Marks: +1, -0

{\left( {\frac{{1 + i}}{{1 - i}}} \right)^n}

= 1?

Solution:

Concept:

Simplify the complex fraction to

i

, then find the smallest positive

n

such that

i^n = 1

Explanation:

We are given

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

First, simplify the base by multiplying numerator and denominator by the conjugate

(1+i)

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

Expand:

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

(since

i^2 = -1

Denominator:

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

Thus

\frac{2i}{2} = i

So the expression becomes

i^n = 1

Recall powers of

i

i^1 = i

i^2 = -1

i^3 = -i

i^4 = 1

The smallest positive integer

n

for which

i^n = 1

n = 4

Answer:

4

(option B).

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