Question

If ${\left(\frac{1+i}{1-i}\right)}^m=1,$ then the least positive integral value of $m$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

${\left(\frac{1+i}{1-i}\right)}^m={(\frac{1+i}{1-i}\times \frac{1+i}{1+i})}^m$
$=(\frac{2i}{2}{)}^m$
$\Rightarrow i^m=1$
$i^2=-1,\Rightarrow i^4=1$
So, the least value of $m$ is $4$