If $(1 - i)^{n} = 2^{n}, t h e n n =$
[RPET 1990]

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Solution

The correct option is B
0

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
$| (1 - i)^{n} | = | 2^{n} | \Rightarrow | 1 - i |^{n} = | 2 |^{n}, (∵ 2^{n} > 0)$
$\Rightarrow [\sqrt{1^{2} + (- 1)^{2}}] = 2^{n} \Rightarrow (\sqrt{2})^{n} = 2^{n}$
$2^{n / 2} = 2^{n} \Rightarrow \frac{n}{2} = n \Rightarrow n = 0$

Trick : By inspection, $(1 - i)^{0} = 2^{0} \Rightarrow 1 = 1$

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If (1−i)n=2n,thenn= [RPET 1990]

If $(1 - i)^{n} = 2^{n}, t h e n n =$
[RPET 1990]