JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Integral power of iota, Algebraic operations and Equality of complex numbers

  • question_answer If  \[{{(1-i)}^{n}}={{2}^{n}},\]then \[n=\] [RPET 1990]

    A) 1

    B) 0

    C) \[-1\]

    D) None of these

    Correct Answer: B

    Solution :

    If \[{{(1-i)}^{n}}={{2}^{n}}\]                   ......(i) We know that if two complex numbers are equal, their moduli must also be equal, therefore from (i), we have \[|{{(1-i)}^{n}}|\,=\,|{{2}^{n}}|\]\[\Rightarrow \]\[|1-i{{|}^{n}}=\,|2{{|}^{n}}\],     \[(\because \,\,{{2}^{n}}>0)\] Þ \[{{\left[ \sqrt{{{1}^{2}}+{{(-1)}^{2}}} \right]}^{n}}={{2}^{n}}\]Þ \[{{(\sqrt{2})}^{n}}={{2}^{n}}\] Þ  \[{{2}^{n/2}}={{2}^{n}}\]Þ \[\frac{n}{2}=n\]Þ\[n=0\] Trick: By inspection,  \[{{(1-i)}^{0}}={{2}^{0}}\,\,\,\,\Rightarrow 1=1\]


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