Question

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

Answer 1

${\left(\frac{1+i}{1-i}\right)}^m=1$
${(\frac{1+i}{1-i}\times \frac{1+i}{1+i})}^m=1$
${\left(\frac{(1+i{)}^2}{1^2-i^2}\right)}^m=1$
${\left(\frac{1+2i+i^2}{2}\right)}^m=1$
${\left(\frac{2i}{2}\right)}^m=1$
$(i{)}^m=1\Rightarrow m=4(\because i^4=1)$