1. **Problem:** Calculate the product $i(3 + 4i)$. 2. **Recall:** Multiplying complex numbers involves distributing and using $i^2 = -1$. 3. Multiply: $$i(3 + 4i) = i \cdot 3 + i \cdot 4i = 3i + 4i^2$$ 4. Substitute $i^2 = -1$: $$3i + 4(-1) = 3i - 4$$ 5. Write in standard form $a + bi$: $$-4 + 3i$$ **Final answer:** $-4 + 3i$