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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 :

Option 1)
$2$

Option 2)
$5$

Option 3)
$4$

Option 4)
$3$

Answers (1)

S solutionqc

$x^{2}-2x+2=0$

$x=\frac{2\pm \sqrt{4-8}}{2}=\frac{2\pm \sqrt{-4}}{2}=\frac{2\pm 2i}{2}$

$=1\pm i$

$\alpha =1+i$

$\beta =1-i$ $\left ( \frac{\alpha }{\beta } \right )=\left ( \frac{1+i}{1-i} \right )=\left ( \frac{(1+i) (1+i)}{(1+i)(1-i)} \right )$

$=\left ( \frac{(1+i)^{2}}{1+1} \right )=\frac{(1+i)^{2}}{2}$

$=\frac{1-1+2i}{2}=i$

So $(i)^{n}=1\Rightarrow i=4$

Option 1)

$2$

Option 2)

$5$

Option 3)

$4$

Option 4)

$3$

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