\( \left. \begin{array} { | l | } \hline \text { Operations on Complex Numbers } \\ \hline \text { Simplify each of the following and write in the form a + bi. } \\ \hline ( 8 + 6 i ) - ( 9 + 3 i ) = \\ \hline ( 11 - i ) ( 15 + i ) = \\ \hline \end{array} \right. \)

Simplify the expression by following steps: - step0: Solution: \(\left(8+6i\right)-\left(9+3i\right)\) - step1: Remove the parentheses: \(8+6i-\left(9+3i\right)\) - step2: Remove the parentheses: \(8+6i-9-3i\) - step3: Add the numbers: \(-1+\left(6-3\right)i\) - step4: Calculate: \(-1+3i\) Expand the expression \( (11 - i)(15 + i) \) Simplify the expression by following steps: - step0: Multiply the numbers: \(\left(11-i\right)\left(15+i\right)\) - step1: Apply the distributive property: \(11\times 15+11i-i\times 15-i\times i\) - step2: Multiply the numbers: \(165+11i-i\times 15-i\times i\) - step3: Multiply the numbers: \(165+11i-15i-i\times i\) - step4: Multiply the numbers: \(165+11i-15i+1\) - step5: Calculate: \(166+11i-15i\) - step6: Calculate: \(166-4i\) Let's simplify the given complex number operations step by step. 1. **For the expression** \( (8 + 6i) - (9 + 3i) \): - We subtract the real parts and the imaginary parts separately: - Real part: \( 8 - 9 = -1 \) - Imaginary part: \( 6 - 3 = 3 \) - Therefore, the result is: \[ -1 + 3i \] 2. **For the expression** \( (11 - i)(15 + i) \): - We will use the distributive property (also known as the FOIL method for binomials): - First: \( 11 \cdot 15 = 165 \) - Outer: \( 11 \cdot i = 11i \) - Inner: \( -i \cdot 15 = -15i \) - Last: \( -i \cdot i = -i^2 = 1 \) (since \( i^2 = -1 \)) - Now, combine all these results: \[ 165 + 11i - 15i + 1 = 166 - 4i \] Thus, the simplified results are: - For \( (8 + 6i) - (9 + 3i) \): \( -1 + 3i \) - For \( (11 - i)(15 + i) \): \( 166 - 4i \)