1. The problem is to simplify the expression $ (1 + i)(2 + 3i)(4 - 3i) $ where $i$ is the imaginary unit with $i^2 = -1$. 2. First, multiply the first two complex numbers: $$ (1 + i)(2 + 3i) = 1 \cdot 2 + 1 \cdot 3i + i \cdot 2 + i \cdot 3i = 2 + 3i + 2i + 3i^2 $$ 3. Since $i^2 = -1$, substitute: $$ 2 + 3i + 2i + 3(-1) = 2 + 5i - 3 = (2 - 3) + 5i = -1 + 5i $$ 4. Now multiply this result by the third complex number: $$ (-1 + 5i)(4 - 3i) = (-1) \cdot 4 + (-1)(-3i) + 5i \cdot 4 + 5i(-3i) = -4 + 3i + 20i - 15i^2 $$ 5. Substitute $i^2 = -1$ again: $$ -4 + 3i + 20i - 15(-1) = -4 + 23i + 15 = (-4 + 15) + 23i = 11 + 23i $$ 6. The simplified form of the expression is $\boxed{11 + 23i}$.