Question

If ${\left(\frac{1+i}{1-i}\right)}^n=-1,/n\in N$, then least value of $n$ is ?

• A. $-1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $-1$

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

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

AS $N\in N$ when $n=1$

${\left[i\right]}^1=i$

When $n=2$

${\left[i\right]}^2=i^2=-1$

$\therefore$ the least value of $n$ such that ${\left(\frac{1+i}{1-i}\right)}^n=-1$ is $2$