• # question_answer If  ${{(1-i)}^{n}}={{2}^{n}},$then $n=$ [RPET 1990] A) 1 B) 0 C) $-1$ D) None of these

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,  ${{(1-i)}^{0}}={{2}^{0}}\,\,\,\,\Rightarrow 1=1$