Question:
If $\alpha$ and $\beta$ be the roots of the equation $x^{2}-2 x+2=0$, then the least value of $n$ for which $\left(\frac{\alpha}{\beta}\right)^{n}=1$ is :
Correct Option: , 3
Solution:
The given quadratic equation is $x^{2}-2 x+2=0$
Then, the roots of the this equation are $\frac{2 \pm \sqrt{-4}}{2}=1 \pm i$
Now, $\frac{\alpha}{\beta}=\frac{1-i}{1+i}=\frac{(1-i)^{2}}{1-i^{2}}=i$
or $\frac{\alpha}{\beta}=\frac{1-i}{1+i}=\frac{(1-i)^{2}}{1-i^{2}}=i$
So, $\frac{\alpha}{\beta}=\pm i$
Now, $\left(\frac{\alpha}{\beta}\right)^{n}=1 \Rightarrow(\pm \mathrm{i})^{n}=1$
$\Rightarrow n$ must be a multiple of 4
Hence, the required least value of $n=4$.