The smallest positive integer n for which

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

where i = √-1, is

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Answer (Detailed Solution Below)

Option 1 : 2
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Detailed Solution

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Concept:

(i)2 = 1

(-i)4 = 1

Calculation:

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

On rationalizing,

\(\rm \left (\frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} \right )^{n^{2}}\)= 1

\(\rm \left (\frac{(1 - i)^{2}}{1 - i^{2}} \right )^{n^{2}}\) = 1

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

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

\(\rm \left (\frac{- 2i}{2} \right )^{n^{2}}\)= 1

\(\rm \left (- i \right )^{n^{2}} \) = 1

if we put n = 2 then, 

\(\rm \left (- i \right )^{n^{2}} \) = 1

Satisfy the equation.

Hence, Option 1 is correct.

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