Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

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

Rationalising the L.H.S,

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

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

$\Rightarrow i^n=1$

$n=4$, smallest position integer