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प्रश्न
Find the smallest positive integer value of m for which \[\frac{(1 + i )^n}{(1 - i )^{n - 2}}\] is a real number.
उत्तर
\[\frac{\left( 1 + i \right)^m}{\left( 1 - i \right)^{m - 2}}\]
\[ = \frac{\left( 1 + i \right)^m}{\left( 1 - i \right)^m} \times \left( 1 - i \right)^2 \]
\[ = \left( \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} \right)^m \times \left( 1 + i^2 - 2i \right)\]
\[ = \left( \frac{1 + i^2 + 2i}{1 - i^2} \right)^m \times \left( 1 - 1 - 2i \right)\]
\[ = \left( \frac{1 - 1 + 2i}{1 + 1} \right)^m \times \left( - 2i \right)\]
\[ = - 2i\left( i^m \right)\]
\[ = - 2 \left( i \right)^{m + 1} \]
\[\text { For this to be real, the smallest positive value of m will be }1 . \]
\[\text{Thus}, i^{1 + 1} = i^2 = - 1,\text { which is real } .\]
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