1. **State the problem:** Simplify the expression $$(1+i)(2+3i)(4-3i)$$ where $i$ is the imaginary unit with the property $i^2 = -1$. 2. **Recall the formula and rules:** When multiplying complex numbers, use the distributive property (FOIL) and remember that $i^2 = -1$. 3. **Step 1: Multiply the first two factors:** $$(1+i)(2+3i) = 1\cdot2 + 1\cdot3i + i\cdot2 + i\cdot3i = 2 + 3i + 2i + 3i^2$$ Since $i^2 = -1$, substitute: $$2 + 3i + 2i + 3(-1) = 2 + 5i - 3 = (2 - 3) + 5i = -1 + 5i$$ 4. **Step 2: Multiply the result by the third factor:** $$(-1 + 5i)(4 - 3i) = (-1)\cdot4 + (-1)(-3i) + 5i\cdot4 + 5i(-3i) = -4 + 3i + 20i - 15i^2$$ Again, substitute $i^2 = -1$: $$-4 + 3i + 20i - 15(-1) = -4 + 23i + 15 = (-4 + 15) + 23i = 11 + 23i$$ 5. **Final answer:** $$\boxed{11 + 23i}$$ This is the simplified form of the given expression.