Question

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

Answer 1

For n = 1, we have,

${\left(\frac{1+i}{1-i}\right)}^n=\frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{(1+i{)}^2}{1^2+1^2}=\frac{1^2+i^2+2\times 1\times i}{2}=\frac{2i}{2}(\because i^2=-1)=i,\text{which is not real}Forn=2,\text{we have}{\left(\frac{1+i}{1-i}\right)}^2=i^2(\because \frac{1+i}{1-i}=1\text{form above})=-1,\text{which is real}$

Hence the least positive integral value of n is 2.