The given expression is:

\[
(-2 + \sqrt{3}i) \times (-2 - \sqrt{3}i)
\]

This is a product of two conjugates, which follows the difference of squares formula:

\[
(a + bi)(a - bi) = a^2 - (bi)^2
\]

For \((-2 + \sqrt{3}i)\) and \((-2 - \sqrt{3}i)\), we have:
- \(a = -2\)
- \(b = \sqrt{3}\)

Now, applying the formula:

\[
a^2 - (bi)^2 = (-2)^2 - (\sqrt{3}i)^2
\]

Calculating each part:
- \((-2)^2 = 4\)
- \((\sqrt{3}i)^2 = (\sqrt{3})^2 \times i^2 = 3 \times (-1) = -3\)

Thus:

\[
4 - (-3) = 4 + 3 = 7
\]

So, the result is:

\[
7
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you simplify the product of complex conjugates?
2. What is the difference of squares rule, and how does it apply to complex numbers?
3. How do you multiply two complex numbers in general form?
4. What does it mean for two numbers to be conjugates in the complex plane?
5. Can you explain why \(i^2 = -1\) and its significance in complex multiplication?

**Tip:** The product of complex conjugates always results in a real number!

Learn how to simplify the product of complex conjugates (-2 + √3i) * (-2 - √3i) using the difference of squares formula. This problem covers key concepts in complex numbers, including conjugates and the difference of squares. Suitable for students in Grades 10-12.

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