Question

The smallest integer n such that ${\left(\frac{1+i}{1-i}\right)}^n=1$ is

• A. 16
• B. 12
• C. 8
• D. 4

Answer 1

The correct option is D 4

${\left(\frac{1+i}{1-i}\right)}^n=1$ $\{\because -i=\Rightarrow \frac{1}{i}\}$
$\Rightarrow {\left(\frac{1+i}{1+\frac{1}{i}}\right)}^n=1$
$i^n=1$
so min value of $n=4$