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UPSC NDA Math Model Paper 7

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Question : 17 of 120

Marks: +1, -0

What

\frac{(1+i)^{4n+5}}{(1-i)^{4n+3}}

is equal to, where

n

is a natural number and

=\sqrt{-1}

?

2
2i
- 2i
i

Solution:

\frac{(1+i)^{4n+5}}{(1-i)^{4n+3}}=\frac{(1+i)^{4n+3} \cdot (1+i)^{2}}{(1-i)^{4n+3}}

=\left(\frac{1+i}{1-i}\right)^{4n+3} \cdot (1+i)^{2}

Rationalizing numerator and denominator by

1+i

, we get

=\left[\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right]^{4n+3} \cdot (1+i^{2}+2i)

=\left[\frac{1+i^{2}+2i}{1+1}\right]^{4n+3} \cdot 2i

=i^{4n+3} \cdot 2i = 2(i)^{4n+4} = 2

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