Question

If $\alpha$ and $\beta$ be the roots of the equation $x^2-2x+2=0$, then the least value of n for which ${\left(\frac{\alpha }{\beta }\right)}^n=1$ is :

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

$x^2-2x+1+1=(x-1{)}^2+1=0\Rightarrow (x-1{)}^2=i^2\Rightarrow (x-1)=\pm i$

$x=(i+1)or(-i+1)$

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

$\therefore$ n (least natural number $=4$