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Find the Smallest Positive Integer Value of M for Which ( 1 + I ) N ( 1 − I ) N − 2 is a Real Number. - Mathematics

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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.

 
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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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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 11 | पृष्ठ ३२

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