Question

If $\alpha$ and $\beta$are the roots of $x^2-2x+2=0$ then the minimum value of $n$ such that ${(\alpha /\beta )}^n=1$

• A. $4$
• B. $3$
• C. $2$
• D. $5$

Answer 1

The correct option is A

$4$

Explanation for correct option:

Step 1. Find the value of $\alpha$ and $\beta$:

Given that $\alpha$ and $\beta$are the roots of $x^2-2x+2=0$.

$$\begin{gathered} x^2-2x+2=0 \\ \Rightarrow x^2-2x+1+1=0 \\ \Rightarrow {(x-1)}^2=-1 \\ \Rightarrow (x-1)=\pm i \\ \Rightarrow x=1\pm i \end{gathered}$$

Let $\alpha =1+i$ then $\beta =1-i$

Step 2. Find the minimum value of $n$:

$$\begin{gathered} \left(\frac{\alpha }{\beta }\right)=\left(\frac{1+i}{1-i}\right) \\ \Rightarrow \left(\frac{\alpha }{\beta }\right)=\frac{{(1+i)}^2}{2} \\ \Rightarrow \left(\frac{\alpha }{\beta }\right)=\frac{(1+i^2+2i)}{2} \\ \Rightarrow \left(\frac{\alpha }{\beta }\right)=\frac{(1-1+2i)}{2} \\ \Rightarrow \left(\frac{\alpha }{\beta }\right)=\frac{2i}{2} \\ \Rightarrow \left(\frac{\alpha }{\beta }\right)=i \end{gathered}$$

Now $$\begin{gathered} {\left(\frac{\alpha }{\beta }\right)}^n=1 \\ {\left(i\right)}^n=1 \end{gathered}$$

Least value of $n$ is $4$.

Hence the correct option is A.