Question

$\text{The smallest positive integer n for which}{\left(\frac{1+i}{1-i}\right)}^n=1\text{is\_\_\_\_\_.}$

• A. 4

Answer 1

The correct option is A 4
${\left(\frac{1+i}{1-i}\right)}^n=1$

$\Rightarrow {(\frac{1+i}{1-i}\times \frac{1+i}{1+i})}^n=1$

$\Rightarrow {\left(\frac{(1+i{)}^2}{1-i^2}\right)}^n=1$

$\Rightarrow {\left(\frac{1+i^2+2i}{1+1}\right)}^n=1$

$\Rightarrow {\left(\frac{1-1+2i}{2}\right)}^n=1$

$\Rightarrow i^n=1$
$\Rightarrow n=4$