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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