The smallest positive integer n for which (1 | vedclass

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(B) Given the equation $(1 + i)^{2n} = (1 - i)^{2n}$.Dividing both sides by $(1 - i)^{2n}$,we get $\left( \frac{1 + i}{1 - i} \right)^{2n} = 1$.Simpli

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$2$

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(B) Given the equation $(1 + i)^{2n} = (1 - i)^{2n}$.Dividing both sides by $(1 - i)^{2n}$,we get $\left( \frac{1 + i}{1 - i} ight)^{2n} = 1$.Simplifying the fraction $\frac{1 + i}{1 - i}$ by multiplying the numerator and denominator by $(1 + i)$:$\frac{1 + i}{1 - i} 	imes \frac{1 + i}{1 + i} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i$.Substituting this back into the equation,we get $i^{2n} = 1$.We know that $i^k = 1$ when $k$ is a multiple of $4$.Thus,$2n = 4$,which gives $n = 2$.

The smallest positive integer $n$ for which $(1 + i)^{2n} = (1 - i)^{2n}$ is

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