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

The correct option is B

8

$\text{Let}z={\left(\frac{2i}{1+i}\right)}^n\Rightarrow z=\frac{2i}{1+i}\times \frac{1-i}{1-i}\Rightarrow z=\frac{2i(1-i)}{1-i^2}\Rightarrow z=\frac{2i(1-i)}{1+1}[\because i^2=-1]\Rightarrow z=\frac{2i(1-i)}{2}\Rightarrow z=i-i^2\Rightarrow z=i+1\text{Now,}{z}^n=(1+i{)}^n\text{For}n=2,z^2=(1+i{)}^2=1+i^2+2i=1-1+2i=2i...\left(1\right)$

Since this is not a positive integer,

$\text{For n =4},z^4=(1+i{)}^4=[(1+i{)}^2{]}^2=(2i{)}^2\left[\text{Using (1)}\right]=4i^2=-4...\left(2\right)$

This is a negative integer,

For n = 8,

$z^8=(1+i{)}^8=[(1+i{)}^4{]}^2=(-4{)}^2\left[\text{Using (2)}\right]=16$

This is a positive integer,

This, $z={\left(\frac{2i}{1+i}\right)}^n$ is positive for n = 8

Therefore, 8 is the least positive integer such that ${\left(\frac{2i}{1+i}\right)}^n$ is a positive integer.