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 can use the quadratic formula to find the roots of the equation \( ax^2 + bx + c = 0 \):

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

For our equation, \( a = 1, b = -2, c = 2 \).

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 we calculate the roots:

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

Thus, we have:

\[ \alpha = 1 + i, \quad \beta = 1 - i. \]

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

Now, we find \( \frac{\alpha}{\beta} \):

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

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

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

### Step 3: Set up the equation \( \left(\frac{\alpha}{\beta}\right)^n = 1 \)

Now we have:

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

### Step 4: Determine when \( i^n = 1 \)

The powers of \( i \) cycle every 4:

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

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

### Conclusion

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

\[ \boxed{4}. \]

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