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

Answer: (C)

${\left(x-1\right)}^2+1=0\Rightarrow x=1+i,1-i$
$\therefore {\left(\frac{\alpha }{\beta }\right)}^n=1\Rightarrow {\left(\pm i\right)}^n=1$
$\therefore$ $n$ (least natural number) = $4$