Question

The smallest positive integral value of n for which ${\left[\frac{1-i}{1+i}\right]}^n$ is purely imaginary with positive imaginary part is

• A. 1
• B. 3
• C. 5
• D. none of thees

Answer 1

Answer: (B)

3

Let, $z={\left[\frac{1-i}{1+i}\right]}^n$
$\Rightarrow z={[\frac{1-i}{1+i}\cdot \frac{1-i}{1-i}]}^n={\left(\frac{1-2i+i^2}{1-i^2}\right)}^n={\left(\frac{-2i}{2}\right)}^n[\because i^2=-1]=(-i{)}^n$
Thus for $z$ to be purely imaginary with positive imaginary part
minimum positive integral value of $n$ is $3,z=(-i{)}^3=-i^3=-i(i^2)=-i(-1)=i$