Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$\text{Given:}$

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

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

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

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

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

$2^n{i}^n=2^n$

$i^n=1$

$i^n=i^4$

$n=4$

option $D$ is correct.