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