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$I f α a n d β b e t h e r o o t s o f t h e e q u a t i o n x^{2} - 2 x + 2 = 0, t h e n t h e l e a s t v a l u e o f n f o r w h i c h {(\frac{α}{β})}^{n} = 1 i s$

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a

3

b

4

c

2

d

5

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n x^{2} - 2 x + 2 = 0 \\ x = \frac{2 \pm \sqrt{4 - 4 (2)}}{2} = 1 \pm i \\ ∴ α = 1 + i, β = 1 - i \\ {(\frac{α}{β})}^{n} = {(\frac{1 + i}{1 - i})}^{n} = {(\frac{{(1 + i)}^{2}}{1 + 1})}^{n} = i^{n} \\ I f n = 4 t h e n i^{4} = 1 \end{array}$

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