1. Find the roots of the quadratic

   Given: $x^2 - 2x + 2 = 0$

   Using the quadratic formula, $x = \frac{2 \pm \sqrt{(-2)^2 - 4\cdot 1 \cdot 2}}{2} = \frac{2 \pm \sqrt{4-8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i$

   So the roots are: $\alpha = 1+i, \quad \beta = 1-i$

2. Compute $\dfrac{\alpha}{\beta}$

   $\frac{\alpha}{\beta} = \frac{1+i}{1-i}$

   Rationalizing: $\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{1-(-1)} = \frac{1+2i+i^2}{2} = \frac{1+2i-1}{2} = \frac{2i}{2} = i$

   Hence, $\frac{\alpha}{\beta} = i$

3. Find the least $n$ such that $\left(\frac{\alpha}{\beta}\right)^n = 1$

   Since $\dfrac{\alpha}{\beta} = i$, we need: $i^n = 1$

   Powers of $i$ repeat every 4: $i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$

   Therefore, the least positive integer $n$ is: $n=4$

4. Check options

   • A: $2$ gives $i^2=-1 \ne 1$
   • B: $5$ gives $i^5=i \ne 1$
   • C: $4$ gives $i^4=1$ ✅
   • D: $3$ gives $i^3=-i \ne 1$

So the correct option is C.