1. **State the problem:** We are given a cubic equation $$z^3 + a z^2 + b z - 5 = 0$$ and one root is $$2 + j$$, where $$j$$ is the imaginary unit. We need to find the values of $$a$$ and $$b$$. 2. **Important rule:** For polynomials with real coefficients, complex roots come in conjugate pairs. Since $$2 + j$$ is a root, its conjugate $$2 - j$$ is also a root. 3. **Roots:** Let the roots be $$2 + j$$, $$2 - j$$, and $$r$$ (real root). 4. **Form the polynomial from roots:** The polynomial can be expressed as $$$(z - (2 + j))(z - (2 - j))(z - r) = 0$$$ 5. **Multiply the first two factors:** $$$(z - 2 - j)(z - 2 + j) = ((z - 2) - j)((z - 2) + j) = (z - 2)^2 - j^2 = (z - 2)^2 + 1$$ 6. **Expand:** $$$(z - 2)^2 + 1 = (z^2 - 4z + 4) + 1 = z^2 - 4z + 5$$ 7. **Multiply by the third factor:** $$$(z^2 - 4z + 5)(z - r) = z^3 - r z^2 - 4 z^2 + 4 r z + 5 z - 5 r = z^3 - (r + 4) z^2 + (4 r + 5) z - 5 r$$ 8. **Compare with given polynomial:** $$$z^3 + a z^2 + b z - 5 = z^3 - (r + 4) z^2 + (4 r + 5) z - 5 r$$$ Matching coefficients: - $$a = -(r + 4)$$ - $$b = 4 r + 5$$ - Constant term: $$-5 = -5 r \\ \Rightarrow r = 1$$ 9. **Calculate $$a$$ and $$b$$:** - $$a = -(1 + 4) = -5$$ - $$b = 4(1) + 5 = 9$$ **Final answer:** $$a = -5$$ and $$b = 9$$.