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Find the Least Positive Integral Value of N for Which ( 1 + I 1 − I ) N is Real. - Mathematics

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प्रश्न

Find the least positive integral value of n for which  \[\left( \frac{1 + i}{1 - i} \right)^n\] is real.

उत्तर

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

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

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

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

\[ = \left( \frac{2i}{2} \right)^n \]

\[ = i^n \]

\[\text { For } i^n \text { to be real, the least positive value of n will be 2} . \]

\[\text { As } i^2 = - 1\]

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अध्याय 13: Complex Numbers - Exercise 13.2 [पृष्ठ ३२]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 13 Complex Numbers
Exercise 13.2 | Q 9 | पृष्ठ ३२

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