Question

find least positive integral value of n for which 1+i/1-i to the power n is a real number

Answer 1

$The \: \: answer \: \: is \: \: given \: \: below \\ \\ Now, \: \: \frac{1 + i}{1 - i} \\ \\ = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} \\ \\ = \frac{1 + i + i + {i}^{2} }{1 - {i}^{2} } \\ \\ = \frac{1 + 2i - 1}{1 + 1} \: \: (since \: \: {i}^{2} = - 1) \\ \\ = \frac{2i}{2} \\ \\ = i \\ \\ We \: \: know \: \: that, \: \: \\ {i}^{2} = - 1 \\ and \: \: ( - 1) \: \: is \: \: a \: \: real \: \: number. \\ \\ So, \: \: {i}^{n} = - 1 \: \: gives \: \: n = 2, \\ which \: \: is \: \: the \: \: least \: \: positive \\ integral \: \: value \: \: of \: \: n. \\ \\ Thank \: \: you \: \: for \: \: your \: \: question.$

Answer 2

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