Question:
Write the least positive integral value of $n$ for which $\left(\frac{1+i}{1-i}\right)^{n}$ is real.
Solution:
$\left(\frac{1+i}{1-i}\right)^{n}$
$=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{n}$
$=\left(\frac{1+i^{2}+2 i}{1-i^{2}}\right)^{n}$
$=\left(\frac{1-1+2 i}{1+1}\right)^{n}$
$\Rightarrow\left(\frac{2 i}{2}\right)^{n}$
$=i^{n}$
For $i^{n}$ to be real, the smallest positive value of $n$ will be 2 .
As, $i^{2}=-1$, which is real.
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