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

Using the quadratic formula to solve x^2 - 2x + 2 = 0 yields the complex roots a = 1 + i and b = 1 - i. We convert these roots to polar form, giving a = sqrt(2)(cos(pi/4) + i sin(pi/4)) and b = sqrt(2)(cos(-pi/4) + i sin(-pi/4)). Using De Moivre's theorem, evaluating (a/b)^n is equivalent to raising the ratio to the power n, which simplifies to cos(n*pi/2) + i sin(n*pi/2). For this expression to equal 1, the angle n*pi/2 must be the smallest positive multiple of 2pi, which occurs when n = 4.

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