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The least positive integer $n$ for which $(1+i)^n=(1-i)^n$, is
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Verified Answer
The correct answer is:
$4$
Given, $(1+i)^n=(1-i)^n$
$$
\begin{aligned}
& \Rightarrow \quad \frac{(1+i)^n}{(1-i)^n}=1 \\
& \Rightarrow \quad\left[\frac{(1+i) \times(1+i)}{(1-i) \times(1+i)}\right]^n=1 \\
& \Rightarrow \quad\left[\frac{1^2+i^2+2 i}{1^2-i^2}\right]^n=1 \\
& \Rightarrow \quad\left[\frac{1-1+2 i}{1+1}\right]^n=1 \\
& \Rightarrow \quad\left(\frac{2 i}{2}\right)^n=1 \\
& \Rightarrow \quad(i)^n=1 \\
& \therefore \quad n=4 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad \frac{(1+i)^n}{(1-i)^n}=1 \\
& \Rightarrow \quad\left[\frac{(1+i) \times(1+i)}{(1-i) \times(1+i)}\right]^n=1 \\
& \Rightarrow \quad\left[\frac{1^2+i^2+2 i}{1^2-i^2}\right]^n=1 \\
& \Rightarrow \quad\left[\frac{1-1+2 i}{1+1}\right]^n=1 \\
& \Rightarrow \quad\left(\frac{2 i}{2}\right)^n=1 \\
& \Rightarrow \quad(i)^n=1 \\
& \therefore \quad n=4 \\
&
\end{aligned}
$$
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