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

n = 2

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n = 0, 2

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n = 0

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n = 1

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Solution

The correct option is C $n = 0$

Given, $(1 - i)^{n} = 2^{n}$
$∴ {\sqrt{2}}^{n} (\frac{1 - i}{\sqrt{2}})^{n} = 2^{n}$
$2^{\frac{n}{2}} . e^{i \frac{n π}{4}} = 2^{n} . e^{i 2 k π}$
Hence, $2 k π = \frac{n π}{4}$
$n = 8 k$ ...(i)
Hence $n$ is of the form 8k.
Also, $2^{\frac{n}{2}} = 2^{n}$
The possible solution is $n = k = 0$ .

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