The integral solution of the equation ${(1 - i)}^{n} = 2^{n}$ is

n = 0

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n = 1

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n = - 1

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None of these

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Solution

The correct option is A $n = 0$
Put $1 = r cos θ$ and $- 1 = r sin θ$

$\Rightarrow r = \sqrt{2}$ and $θ = - \frac{π}{4}$

Then, the given equation takes the form

${(\sqrt{2})}^{n} {[cos (- \frac{π}{4}) + i sin (- \frac{π}{4})]}^{n} = 2^{n}$

$\Rightarrow 2^{n / 2} [cos \frac{n π}{4} - i sin \frac{n π}{4}] = 2^{n}$

Equating real and imaginary parts, we get

$cos \frac{n π}{4} = 2^{n / 2}$ and $sin \frac{n π}{4} = 0$

These are satisfied only for $n = 0$

Hence, $n = 0$ is the only solution.

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