Question

The smallest integral value of $n$ for which ${\left(\frac{1+i}{1-i}\right)}^n=1$ is

• A. $2$
• B. $4$
• C. $6$
• D. none of these

Answer 1

The correct option is B $4$

Now,

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

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

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

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

According to the problem

$i^n=1$.

or, $n=$ multiple of $4$.

smallest ${}^{\prime }{n}^{\prime }$ is $4$.