Question

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

• A. $n=8$
• B. $n=16$
• C. $n=12$
• D. none of these

Answer 1

The correct option is D none of these

Given, ${\frac{(1+i)}{1-i}}^n=1$
$\Rightarrow {\frac{(1+i)(1+i)}{(1-i)(1+i}}^n=1$
$\Rightarrow {\frac{2i}{2}}^n=1$
$i^n=1$
The smallest integral value of $n$ will be $4$.