If `(-2 + sqrt-3) (-3 + 2 sqrt-3) = a + bi`, find the real numbers a and b. With these values of a and b, also find the modulus of `a+ bi`

Answer

To solve the problem, we need to evaluate the expression \((-2 + \sqrt{-3})(-3 + 2\sqrt{-3})\) and express it in the form \(a + bi\), where \(a\) and \(b\) are real numbers. Then, we will find the modulus of the complex number \(a + bi\).

### Step-by-Step Solution:

1. **Rewrite the Square Roots**:
We know that \(\sqrt{-3} = \sqrt{3}i\) (where \(i = \sqrt{-1}\)). Thus, we can rewrite the expression:
\[ (-2 + \sqrt{-3})(-3 + 2\sqrt{-3}) = (-2 + \sqrt{3}i)(-3 + 2\sqrt{3}i) \]

2. **Expand the Expression**:
We will use the distributive property (FOIL method) to expand the expression:
\[ = (-2)(-3) + (-2)(2\sqrt{3}i) + (\sqrt{3}i)(-3) + (\sqrt{3}i)(2\sqrt{3}i) \]
Calculating each term:
- \((-2)(-3) = 6\)
- \((-2)(2\sqrt{3}i) = -4\sqrt{3}i\)
- \((\sqrt{3}i)(-3) = -3\sqrt{3}i\)
- \((\sqrt{3}i)(2\sqrt{3}i) = 2(\sqrt{3})^2(i^2) = 2(3)(-1) = -6\)

Combining these results:
\[ = 6 - 4\sqrt{3}i - 3\sqrt{3}i - 6 \]

3. **Combine Like Terms**:
Now, we combine the real and imaginary parts:
\[ = (6 - 6) + (-4\sqrt{3}i - 3\sqrt{3}i) = 0 - 7\sqrt{3}i \]
Thus, we have:
\[ = 0 - 7\sqrt{3}i \]

4. **Identify \(a\) and \(b\)**:
From the expression \(0 - 7\sqrt{3}i\), we can identify:
- \(a = 0\)
- \(b = -7\sqrt{3}\)

5. **Find the Modulus**:
The modulus of a complex number \(a + bi\) is given by the formula:
\[ |a + bi| = \sqrt{a^2 + b^2} \]
Substituting the values of \(a\) and \(b\):
\[ |0 - 7\sqrt{3}i| = \sqrt{0^2 + (-7\sqrt{3})^2} = \sqrt{0 + 49 \cdot 3} = \sqrt{147} = 7\sqrt{3} \]

### Final Result:
- The values of \(a\) and \(b\) are:
- \(a = 0\)
- \(b = -7\sqrt{3}\)
- The modulus of \(a + bi\) is:
- \(|a + bi| = 7\sqrt{3}\)