Question

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

Answer 1

$$\begin{gathered} {\left(\frac{1+i}{1-i}\right)}^n \\ ={\left[\frac{1+i}{1-i}\times \left(\frac{1+i}{1+i}\right)\right]}^n \\ ={\left(\frac{1+i^2+2i}{1-i^2}\right)}^n \\ ={\left(\frac{1-1+2i}{1+1}\right)}^n \\ ={\left(\frac{2i}{2}\right)}^n \\ =i^n \\ \mathrm{For}{i}^n\mathrm{to}\mathrm{be}\mathrm{real},\mathrm{the}\mathrm{least}\mathrm{positive}\mathrm{value}\mathrm{of}n\mathrm{will}\mathrm{be}2. \\ \mathrm{As}{i}^2=-1 \end{gathered}$$