To solve the given complex multiplication problem:

$(2 + j)(3 - j^2) = a + jb$

Step-by-Step Solution:

1. Simplify $j^2$: $j^2 = -1$ So the expression becomes: $(2 + j)(3 - (-1)) = (2 + j)(3 + 1) = (2 + j)(4)$

2. Distribute the terms: Now, distribute $4$ to both parts of $(2 + j)$: $4 \cdot (2 + j) = 4 \cdot 2 + 4 \cdot j = 8 + 4j$

Thus, the expression simplifies to: $8 + 4j = a + jb$

Conclusion:

$a = 8, \quad b = 4$

So, the solution is: $(2 + j)(3 - j^2) = 8 + 4j$

Do you have any questions or would you like more details?

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Here are 5 related questions for practice:

1. What is the result of $(1 + 2j)(3 - j)$?
2. Simplify $(2 - j)(2 + j)$.
3. What is $(3 + 2j)^2$?
4. If $j = \sqrt{-1}$, what is $j^3$?
5. Find the product $(1 + j)(1 - j)$.

Tip: When multiplying complex numbers, remember to distribute terms and apply the rule $j^2 = -1$ to simplify.