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Step-by-Step Solution
Step 1: Identify the Quadratic Equation
The given equation is
$ x^2 - 2x + 2 = 0. $
We will find its roots using the quadratic formula.
Step 2: Compute the Roots
Using the quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, $
where $a = 1,\, b = -2,\, c = 2,$ we get
\[
x \;=\; \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2}
\;=\; \frac{2 \pm \sqrt{4 - 8}}{2}
\;=\; \frac{2 \pm \sqrt{-4}}{2}
\;=\; 1 \pm i.
\]
Thus, the roots are
$ \alpha = 1 + i \; \text{and} \; \beta = 1 - i. $
Step 3: Form the Ratio $ \alpha / \beta $
Compute the ratio
\[
\frac{\alpha}{\beta} = \frac{1 + i}{1 - i}.
\]
To simplify, multiply numerator and denominator by $ (1 - i) $:
\[
\frac{\alpha}{\beta}
= \frac{(1 + i)(1 - i)}{(1 - i)(1 - i)}
= \frac{(1 + i)(1 - i)}{(1)^2 - (i)^2}
= \frac{1 - i^2}{1 + 1}
= \frac{1 - (-1)}{2}
= \frac{2}{2}
= 1 \, \text{(incorrect if carried differently, see below for correct approach)}.
\]
However, the known result (and more direct method) shows:
\[
\frac{(1 + i)}{(1 - i)}
= \frac{(1 + i)^2}{(1 - i)(1 + i)}
= \frac{1 + 2i + i^2}{1 - i^2}
= \frac{1 + 2i - 1}{1 - (-1)}
= \frac{2i}{2}
= i.
\]
So, one possibility is:
\[
\frac{\alpha}{\beta} = i.
\]
Alternatively, if $ \alpha = 1 - i $ and $ \beta = 1 + i, $ we get
\[
\frac{\alpha}{\beta} = \frac{1 - i}{1 + i} = -\,i.
\]
Step 4: Find Condition for $ \bigl(\frac{\alpha}{\beta}\bigr)^n = 1 $
Since $ \frac{\alpha}{\beta} $ is either $ i $ or $ -i, $ we check:
\[
(i)^n = 1 \quad \text{and} \quad (-i)^n = 1.
\]
We know from properties of complex numbers that $ i^4 = 1. $ Hence, $ i^n = 1 $ if and only if $ n $ is a multiple of 4. The same argument applies to $ (-i)^n. $
Step 5: Conclude the Minimum Positive Value of $ n $
The smallest positive integer multiple of 4 is 4 itself. Hence, the least value of $ n $ for which
$ \left(\frac{\alpha}{\beta}\right)^n = 1 $
is
\[
\boxed{4}.
\]
