1. **Problem:** Find the value of (a) $(3+2i)(3-3i)$. 2. **Formula and rules:** To multiply complex numbers, use the distributive property and remember that $i^2 = -1$. 3. **Intermediate work:** $$(3+2i)(3-3i) = 3\times3 + 3\times(-3i) + 2i\times3 + 2i\times(-3i)$$ $$= 9 - 9i + 6i - 6i^2$$ $$= 9 - 3i - 6(-1)$$ $$= 9 - 3i + 6$$ $$= 15 - 3i$$ 4. **Explanation:** We multiplied each term and combined like terms. Since $i^2 = -1$, $-6i^2$ becomes $+6$. 5. **Final answer:** The value of $(3+2i)(3-3i)$ is $15 - 3i$.