If a and b are real numbers between 0 and 1 such that the points ` z_(1) = a+ i , z_(2) = 1 + bi and z_(3) = 0 ` from an equilateral triangle then (a,b) =

Options

A. `(2 - sqrt(3) , 2 - sqrt(3))`
B. ` (1 // 3, 1//3)`
C. `(1//2,1//2)`
D. none

Answer

To solve the problem, we need to find the values of \(a\) and \(b\) such that the points \(z_1 = a + i\), \(z_2 = 1 + bi\), and \(z_3 = 0\) form an equilateral triangle in the complex plane.

### Step 1: Calculate the distances between the points
The distances between the points \(z_1\), \(z_2\), and \(z_3\) must be equal since they form an equilateral triangle.

1. **Distance \(z_1\) to \(z_3\)**:
\[ |z_1 - z_3| = |(a + i) - 0| = \sqrt{a^2 + 1^2} = \sqrt{a^2 + 1} \]

2. **Distance \(z_2\) to \(z_3\)**:
\[ |z_2 - z_3| = |(1 + bi) - 0| = \sqrt{1^2 + b^2} = \sqrt{1 + b^2} \]

3. **Distance \(z_1\) to \(z_2\)**:
\[ |z_1 - z_2| = |(a + i) - (1 + bi)| = |(a - 1) + (1 - b)i| = \sqrt{(a - 1)^2 + (1 - b)^2} \]

### Step 2: Set the distances equal
Since the distances are equal, we can set up the following equations:
1. \( \sqrt{a^2 + 1} = \sqrt{1 + b^2} \)
2. \( \sqrt{1 + b^2} = \sqrt{(a - 1)^2 + (1 - b)^2} \)

### Step 3: Solve the first equation
Squaring both sides of the first equation:
\[ a^2 + 1 = 1 + b^2 \implies a^2 = b^2 \implies a = b \quad \text{(since \(a\) and \(b\) are both positive)} \]

### Step 4: Substitute \(a = b\) into the second equation
Substituting \(b\) for \(a\) in the second equation:
\[ \sqrt{1 + b^2} = \sqrt{(b - 1)^2 + (1 - b)^2} \]

### Step 5: Simplify the second equation
Squaring both sides:
\[ 1 + b^2 = (b - 1)^2 + (1 - b)^2 \]
Expanding the right side:
\[ (b - 1)^2 + (1 - b)^2 = (b^2 - 2b + 1) + (1 - 2b + b^2) = 2b^2 - 4b + 2 \]
Setting the equations equal:
\[ 1 + b^2 = 2b^2 - 4b + 2 \]
Rearranging gives:
\[ 0 = b^2 - 4b + 1 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula:
\[ b = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]

### Step 7: Determine valid values for \(a\) and \(b\)
Since \(a\) and \(b\) must be between 0 and 1:
1. \(b = 2 + \sqrt{3}\) is not valid (greater than 1).
2. \(b = 2 - \sqrt{3}\) is valid since \(2 - \sqrt{3} \approx 0.268\) (between 0 and 1).

Thus, we have:
\[ (a, b) = (2 - \sqrt{3}, 2 - \sqrt{3}) \]

### Final Answer
\[ (a, b) = (2 - \sqrt{3}, 2 - \sqrt{3}) \]