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Question

(C) Given expression: $Z = \frac{(1 + i)^n}{(1 - i)^{n-2}}$.Multiply numerator and denominator by $(1 + i)^2$ to simplify:$Z = \frac{(1 + i)^n (1 + i)

Vedclass · Accepted Answer

$3$

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(C) Given expression: $Z = \frac{(1 + i)^n}{(1 - i)^{n-2}}$.Multiply numerator and denominator by $(1 + i)^2$ to simplify:$Z = \frac{(1 + i)^n (1 + i)^2}{(1 - i)^{n-2} (1 + i)^2} = \frac{(1 + i)^{n+2}}{((1 - i)(1 + i))^{n-2} (1 + i)^4} = \frac{(1 + i)^{n+2}}{(1^2 - i^2)^{n-2} (1 + i)^4} = \frac{(1 + i)^{n+2}}{2^{n-2} (1 + i)^4} = \frac{(1 + i)^{n-2}}{2^{n-2}} = \left( \frac{1 + i}{2} \right)^{n-2}$.Alternatively, simplify the base: $\frac{1+i}{1-i} = \frac{(1+i)^2}{1^2+1^2} = \frac{1-1+2i}{2} = i$.So, $Z = \frac{(1+i)^n}{(1-i)^{n-2}} = \frac{(1+i)^2 (1+i)^{n-2}}{(1-i)^{n-2}} = (1+i)^2 \left( \frac{1+i}{1-i} \right)^{n-2} = 2i \cdot (i)^{n-2}$.For $Z$ to be a real number, $i \cdot i^{n-2} = i^{n-1}$ must be real.$i^{n-1}$ is real if $n-1$ is a multiple of $2$. The smallest positive integer $n$ such that $n-1 = 2$ is $n = 3$.

The smallest positive integer $n$ for which $\frac{(1 + i)^n}{(1 - i)^{n-2}}$ is a real number, is ...

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