# Find the least positive integral value of n for which

Question:

Find the least positive integral value of $n$ for which $\left(\frac{1+i}{1-i}\right)^{n}$ is real.

Solution:

$\left(\frac{1+i}{1-i}\right)^{n}$

$=\left[\frac{1+i}{1-i} \times\left(\frac{1+i}{1+i}\right)\right]^{n}$

$=\left(\frac{1+i^{2}+2 i}{1-i^{2}}\right)^{n}$

$=\left(\frac{1-1+2 i}{1+1}\right)^{n}$

$=\left(\frac{2 i}{2}\right)^{n}$

$=i^{n}$

For $i^{n}$ to be real, the least positive value of $n$ will be 2 .

As $i^{2}=-1$