Key Concepts and Formulas

• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Complex Numbers: A complex number is of the form $a + bi$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$, such that $i^2 = -1$.
• Complex Conjugate: The complex conjugate of $a + bi$ is $a - bi$.
• Powers of i: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. The powers of $i$ repeat in a cycle of 4.

Step-by-Step Solution

Step 1: Find the roots of the quadratic equation We are given the quadratic equation $x^2 - 2x + 2 = 0$. We will use the quadratic formula to find the roots $\alpha$ and $\beta$. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -2$, and $c = 2$. Substituting these values, we get: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(2)}}{2(1)}$ $x = \frac{2 \pm \sqrt{4 - 8}}{2}$ $x = \frac{2 \pm \sqrt{-4}}{2}$ Since $\sqrt{-4} = \sqrt{4} \cdot \sqrt{-1} = 2i$, we have: $x = \frac{2 \pm 2i}{2}$ $x = 1 \pm i$ Thus, the roots are $\alpha = 1 + i$ and $\beta = 1 - i$.

Step 2: Calculate the ratio $\frac{\alpha}{\beta}$ We will now calculate the ratio $\frac{\alpha}{\beta}$. $\frac{\alpha}{\beta} = \frac{1 + i}{1 - i}$ To simplify this complex fraction, we multiply the numerator and denominator by the complex conjugate of the denominator, which is $1 + i$. $\frac{\alpha}{\beta} = \frac{1 + i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{(1 - i)(1 + i)}$ Expanding the numerator and denominator: $\frac{\alpha}{\beta} = \frac{1 + 2i + i^2}{1 - i^2}$ Since $i^2 = -1$, we have: $\frac{\alpha}{\beta} = \frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i$ If we had chosen $\alpha = 1 - i$ and $\beta = 1 + i$, we would have: $\frac{\alpha}{\beta} = \frac{1 - i}{1 + i} \cdot \frac{1 - i}{1 - i} = \frac{(1 - i)^2}{(1 + i)(1 - i)} = \frac{1 - 2i + i^2}{1 - i^2} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i$ So, $\frac{\alpha}{\beta}$ can be either $i$ or $-i$.

Step 3: Find the least value of $n$ for which $\left(\frac{\alpha}{\beta}\right)^n = 1$

Case 1: $\frac{\alpha}{\beta} = i$ We want to find the smallest positive integer $n$ such that $i^n = 1$. We know the powers of $i$ are cyclic: $i^1 = i$ $i^2 = -1$ $i^3 = -i$ $i^4 = 1$ Thus, the least value of $n$ is $4$.

Case 2: $\frac{\alpha}{\beta} = -i$ We want to find the smallest positive integer $n$ such that $(-i)^n = 1$. We know the powers of $-i$ are cyclic: $(-i)^1 = -i$ $(-i)^2 = -1$ $(-i)^3 = i$ $(-i)^4 = 1$ Thus, the least value of $n$ is $4$.

In both cases, the least value of $n$ for which $\left(\frac{\alpha}{\beta}\right)^n = 1$ is $4$.

Common Mistakes & Tips

• Be careful with signs when using the quadratic formula and simplifying complex numbers.
• Remember to multiply the numerator and denominator by the conjugate of the denominator when dividing complex numbers.
• Memorize the powers of $i$: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.

Summary

We first found the complex roots of the quadratic equation using the quadratic formula. Then, we computed the ratio of the roots, which simplified to either $i$ or $-i$. Finally, we determined the smallest positive integer $n$ such that $\left(\frac{\alpha}{\beta}\right)^n = 1$. In both cases for the ratio, the least value of $n$ is $4$.

Final Answer

The final answer is $\boxed{4}$, which corresponds to option (C).