Question

Find the smallest positive integer value of m for which $\frac{(1+i{)}^n}{(1-i{)}^{n-2}}$ is a real number.

Answer 1

$$\begin{gathered} \frac{{\left(1+i\right)}^m}{{\left(1-i\right)}^{m-2}} \\ =\frac{{\left(1+i\right)}^m}{{\left(1-i\right)}^m}\times {\left(1-i\right)}^2 \\ ={\left(\frac{1+i}{1-i}\times \frac{1+i}{1+i}\right)}^m\times \left(1+i^2-2i\right) \\ ={\left(\frac{1+i^2+2i}{1-i^2}\right)}^m\times \left(1-1-2i\right) \\ ={\left(\frac{1-1+2i}{1+1}\right)}^m\times \left(-2i\right) \\ =-2i\left(i^m\right) \\ =-2{\left(i\right)}^{m+1} \\ \mathrm{For}\mathrm{this}\mathrm{to}\mathrm{be}\mathrm{real},\mathrm{the}\mathrm{smallest}\mathrm{positive}\mathrm{value}\mathrm{of}m\mathrm{will}\mathrm{be}1. \\ \mathrm{Thus},i^{1+1}=i^2=-1,\mathrm{which}\mathrm{is}\mathrm{real}. \end{gathered}$$