If `alpha` and `beta` are the complex roots of the equation `(1+i)x^(2)+(1-i)x-2i=o` where `i=sqrt(-1)`, the value of `|alpha-beta|^(2)` is

Explanation

To solve the equation \((1+i)x^2 + (1-i)x - 2i = 0\) and find the value of \(|\alpha - \beta|^2\) where \(\alpha\) and \(\beta\) are the complex roots, we can follow these steps:

### Step 1: Identify coefficients
The given quadratic equation can be written in the standard form \(ax^2 + bx + c = 0\), where:
- \(a = 1 + i\)
- \(b = 1 - i\)
- \(c = -2i\)

### Step 2: Calculate \(\alpha + \beta\)
Using the formula for the sum of the roots of a quadratic equation:
\[ \alpha + \beta = -\frac{b}{a} = -\frac{1 - i}{1 + i} \]
To simplify this, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \alpha + \beta = -\frac{(1 - i)(1 - i)}{(1 + i)(1 - i)} = -\frac{(1 - 2i + i^2)}{1^2 - i^2} = -\frac{(1 - 2i - 1)}{1 + 1} = -\frac{-2i}{2} = i \]

### Step 3: Calculate \(\alpha \beta\)
Using the formula for the product of the roots:
\[ \alpha \beta = \frac{c}{a} = \frac{-2i}{1 + i} \]
Again, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \alpha \beta = \frac{-2i(1 - i)}{(1 + i)(1 - i)} = \frac{-2i + 2i^2}{1 + 1} = \frac{-2i - 2}{2} = -i - 1 \]

### Step 4: Calculate \(|\alpha - \beta|^2\)
Using the identity:
\[ |\alpha - \beta|^2 = (\alpha + \beta)^2 - 4\alpha \beta \]
Substituting the values we found:
\[ |\alpha - \beta|^2 = (i)^2 - 4(-i - 1) \]
Calculating this:
\[ |\alpha - \beta|^2 = -1 + 4(i + 1) = -1 + 4i + 4 = 3 + 4i \]

### Step 5: Find the modulus
Now, we need to find the modulus of \(3 + 4i\):
\[ |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Thus,
\[ |\alpha - \beta|^2 = 5^2 = 25 \]

### Final Answer
The value of \(|\alpha - \beta|^2\) is \(25\).

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