Question

The least positive integer n such that ${\left(\frac{2i}{1+i}\right)}^n$ is a positive integer, is
(a) 16
(b) 8
(c) 4
(d) 2

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)8 \\ \mathrm{Let}z=\left(\frac{2i}{1+i}\right) \\ \Rightarrow z=\frac{2i}{1+i}\times \frac{1-i}{1-i} \\ \Rightarrow z=\frac{2i\left(1-i\right)}{1-i^2} \\ \Rightarrow z=\frac{2i\left(1-i\right)}{1+1}\left[\because i^2=-1\right] \\ \Rightarrow z=\frac{2i\left(1-i\right)}{2} \\ \Rightarrow z=i-i^2 \\ \Rightarrow z=i+1 \\ \mathrm{Now},z^n={\left(1+i\right)}^n \\ \mathrm{For}n=2, \\ {z}^2={\left(1+i\right)}^2 \\ =1+i^2+2i \\ =1-1+2i \\ =2i...\left(1\right) \\ \mathrm{Since}\mathrm{this}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{positive}\mathrm{integer}, \\ \mathrm{For}n=4, \\ {z}^4={\left(1+i\right)}^4 \\ ={\left[{\left(1+i\right)}^2\right]}^2 \\ ={\left(2i\right)}^2\left[\mathrm{Using}\left(1\right)\right] \\ =4i^2 \\ =-4...\left(2\right) \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{negative}\mathrm{integer}. \\ \mathrm{For}n=8, \\ {z}^8={\left(1+i\right)}^8 \\ ={\left[{\left(1+i\right)}^4\right]}^2 \\ ={\left(-4\right)}^2\left[\mathrm{Using}\left(2\right)\right] \\ =16 \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{integer}. \\ \mathrm{Thus},z={\left(\frac{2i}{1+i}\right)}^n\mathrm{is}\mathrm{positive}\mathrm{for}n=8. \\ \mathrm{Therefore},8\mathrm{is}\mathrm{the}\mathrm{least}\mathrm{positive}\mathrm{integer}\mathrm{such}\mathrm{that}{\left(\frac{2i}{1+i}\right)}^n\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{integer}. \end{gathered}$$