If n is a positive integer,then left( frac{1 | vedclass

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(C) First,simplify the expression inside the parenthesis: $\frac{1 + i}{1 - i} = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - i^2}

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$i$

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(C) First,simplify the expression inside the parenthesis: $\frac{1 + i}{1 - i} = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i$ Now,substitute this into the given expression: $\left( \frac{1 + i}{1 - i} ight)^{4n + 1} = i^{4n + 1}$ Using the properties of exponents: $i^{4n + 1} = i^{4n} 	imes i^1$ Since $i^4 = 1$,it follows that $i^{4n} = (i^4)^n = 1^n = 1$ Therefore,$i^{4n + 1} = 1 	imes i = i$

If $n$ is a positive integer,then $\left( \frac{1 + i}{1 - i} \right)^{4n + 1} = $

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