If alpha and beta are the roots of the equati | vedclass

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Question

(A) The given equation is $x^2 - 2x + 2 = 0$.Using the quadratic formula,$x = \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i$.Let $\alpha = 1 + i$ and $\beta

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The given equation is $x^2 - 2x + 2 = 0$.Using the quadratic formula,$x = \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i$.Let $\alpha = 1 + i$ and $\beta = 1 - i$.Then $\frac{\alpha}{\beta} = \frac{1 + i}{1 - i} = \frac{(1 + i)^2}{(1 - i)(1 + i)} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i$.We need to find the least natural number $n$ such that $(\frac{\alpha}{\beta})^n = 1$,which means $i^n = 1$.The powers of $i$ are $i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1$.Thus,the least value of $n$ is $4$.

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

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