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If $n$ is a positive integer, then $(1+i)^n \div(1-i)^n=-i$, then $n$ will be of the form
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Verified Answer
The correct answer is:
$4 k-1, k \in N$
$\begin{aligned}
& \text { Given, } \frac{(1+i)^n}{(1-i)^n}=-i \\
& \Rightarrow \quad\left(\frac{1+i}{1-i}\right)^n=-i \\
& \Rightarrow \quad\left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^n=-i \\
& \Rightarrow \quad\left[\frac{(1+i)^2}{1+1}\right]^n=-i \\
& \Rightarrow \quad\left[\frac{1-1+2 i}{2}\right]^n=-i \Rightarrow\left[\frac{2 i}{2}\right]^n=-i \\
& \Rightarrow \quad i^n=-i
\end{aligned}$
$i^n=-i$ is possible when $n$ will be of the form $4 k-1, k \in N$.
& \text { Given, } \frac{(1+i)^n}{(1-i)^n}=-i \\
& \Rightarrow \quad\left(\frac{1+i}{1-i}\right)^n=-i \\
& \Rightarrow \quad\left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^n=-i \\
& \Rightarrow \quad\left[\frac{(1+i)^2}{1+1}\right]^n=-i \\
& \Rightarrow \quad\left[\frac{1-1+2 i}{2}\right]^n=-i \Rightarrow\left[\frac{2 i}{2}\right]^n=-i \\
& \Rightarrow \quad i^n=-i
\end{aligned}$
$i^n=-i$ is possible when $n$ will be of the form $4 k-1, k \in N$.
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