Question

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

• A. $5$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is B $4$

$(x-1{)}^2+1=0\Rightarrow x=1\pm i\therefore \frac{\alpha }{\beta }=\left(\frac{1\pm i}{1\mp i}\right)=\pm \left(\frac{2i}{2}\right)=\pm iA.T.Q{\left(\frac{\alpha }{\beta }\right)}^n=(\pm i{)}^n=1\Rightarrow n=4$