If `alpha` and `beta` be the roots of the equation `x^(2)-2x+2=0`, then the least value of `n` for which `((alpha)/(beta))^(n)=1` is:

Options

A. 2
B. 5
C. 4
D. 3

Answer

To solve the problem, we need to find the least value of \( n \) such that

\[ \left(\frac{\alpha}{\beta}\right)^n = 1 \]

where \( \alpha \) and \( \beta \) are the roots of the equation

\[ x^2 - 2x + 2 = 0. \]

### Step 1: Find the roots \( \alpha \) and \( \beta \)

We start by solving the quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = -2, c = 2 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 \]

Since the discriminant is negative, the roots are complex. Now substituting into the quadratic formula:

\[ x = \frac{2 \pm \sqrt{-4}}{2 \cdot 1} = \frac{2 \pm 2i}{2} = 1 \pm i \]

Thus, the roots are:

\[ \alpha = 1 + i \quad \text{and} \quad \beta = 1 - i \]

### Step 2: Calculate \( \frac{\alpha}{\beta} \)

Now we calculate \( \frac{\alpha}{\beta} \):

\[ \frac{\alpha}{\beta} = \frac{1 + i}{1 - i} \]

To simplify this, we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{(1 + 2i - 1)}{1 + 1} = \frac{2i}{2} = i \]

### Step 3: Find \( n \) such that \( \left(\frac{\alpha}{\beta}\right)^n = 1 \)

We need to find the smallest \( n \) such that:

\[ i^n = 1 \]

The powers of \( i \) are:

- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)

Thus, \( i^n = 1 \) when \( n \) is a multiple of 4. The least positive integer \( n \) that satisfies this is:

\[ n = 4 \]

### Conclusion

The least value of \( n \) for which \( \left(\frac{\alpha}{\beta}\right)^n = 1 \) is:

\[ \boxed{4} \]