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( { 1+i }{ 1-i } ) ^{ 402 } is equal to

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$(\frac{1 + i}{1 - i})^{402}$ is equal to

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Text solutionVerified

(\frac{1 + i}{1 - i})^{402}

[\frac{( 1 + i ) ( 1 + i )}{( 1 - i ) ( 1 + i )}]^{402}

on rationalizing denominator,

(\frac{1 ^{2} + i ^{2} + 2 i}{1 ^{2} - i ^{2}})^{402} = (\frac{1 - 1 + 2 i}{1 + 1})^{402}

\Rightarrow (\frac{2 i}{2})^{402} \Rightarrow i^{402} = i^{2} \Rightarrow - 1

Option A is correct

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