Prove the following


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 :

  1. (1) 2

  2. (2) 5

  3. (3) 4

  4. (4) 3

Correct Option: 3,


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 $n=4$.

Leave a comment