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Question
Find the least positive integral value of n for which \[\left( \frac{1 + i}{1 - i} \right)^n\] is real.
Solution
\[\left( \frac{1 + i}{1 - i} \right)^n \]
\[ = \left[ \frac{1 + i}{1 - i} \times \left( \frac{1 + i}{1 + i} \right) \right]^n \]
\[ = \left( \frac{1 + i^2 + 2i}{1 - i^2} \right)^n \]
\[ = \left( \frac{1 - 1 + 2i}{1 + 1} \right)^n \]
\[ = \left( \frac{2i}{2} \right)^n \]
\[ = i^n \]
\[\text { For } i^n \text { to be real, the least positive value of n will be 2} . \]
\[\text { As } i^2 = - 1\]
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