https://file.aiquickdraw.com/mathbot/file/1726197298224fni3ycda.jpg

Let's go through the equations step by step.

### 1. First Equation: \((a + bi)(a - bi)\)

This is a difference of squares problem:
\[
(a + bi)(a - bi) = a^2 - (bi)^2
\]
Since \(i^2 = -1\), we have:
\[
a^2 - (bi)^2 = a^2 - b^2(-1) = a^2 + b^2
\]
Therefore, the answer to the first equation is:
\[
a^2 + b^2 + 0i
\]

### 2. Second Equation: \((a + bi)^2 - (a - bi)^2\)

First, expand both squares:

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

Now, subtract:
\[
(a + bi)^2 - (a - bi)^2 = (a^2 + 2abi - b^2) - (a^2 - 2abi - b^2)
\]
This simplifies to:
\[
= a^2 + 2abi - b^2 - a^2 + 2abi + b^2
\]
The \(a^2\) and \(-b^2\) terms cancel out, leaving:
\[
4abi
\]
So the answer is:
\[
0 + 4abi
\]

### Final Answers:
1. \((a + bi)(a - bi) = a^2 + b^2 + 0i\)
2. \((a + bi)^2 - (a - bi)^2 = 0 + 4abi\)

Would you like more details or have any questions? Here are some related questions:

1. What are the properties of complex conjugates?
2. How can you apply the difference of squares formula to complex numbers?
3. Why does \(i^2 = -1\) play a crucial role in simplifying complex expressions?
4. How would these expressions change if \(b = 0\)?
5. Can you apply this method to multiply any two complex numbers?

**Tip:** Remember that multiplying a complex number by its conjugate always results in a real number!

Complete the following equations involving complex numbers: (a + bi)(a - bi) and (a + bi)^2 - (a - bi)^2.

Learn how to solve complex number equations like (a + bi)(a - bi) and (a + bi)^2 - (a - bi)^2 using algebraic techniques and the difference of squares formula. Suitable for students in Grades 9-12.

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