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!