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) = $4$