1. **State the problem:** Simplify the expression $$(z-(3+i))(z-(3-i))$$. 2. **Recall the formula:** This is a product of two binomials of the form $(z - a)(z - \overline{a})$ where $a = 3+i$ and $\overline{a} = 3 - i$ is its complex conjugate. 3. **Important rule:** The product of a complex number and its conjugate is a real number given by $$|a|^2 = a \times \overline{a}$$. 4. **Expand the expression:** $$ (z-(3+i))(z-(3-i)) = (z-3 - i)(z-3 + i) $$ 5. **Use the difference of squares formula:** $$ (a - b)(a + b) = a^2 - b^2 $$ where $a = z-3$ and $b = i$. 6. **Apply the formula:** $$ (z-3)^2 - i^2 $$ 7. **Simplify $i^2$:** $$ i^2 = -1 $$ 8. **Substitute back:** $$ (z-3)^2 - (-1) = (z-3)^2 + 1 $$ 9. **Final simplified form:** $$ (z-3)^2 + 1 $$ This is the simplified expression of the original product.