1. **Solve the system (i):** Given: $$z - 4w = 3i$$ $$2z + 3w = 11 - 5i$$ Multiply the first equation by 3: $$3z - 12w = 9i$$ Multiply the second by 4: $$8z + 12w = 44 - 20i$$ Add to eliminate $w$: $$11z = 44 - 11i \Rightarrow z = 4 - i$$ Substitute into first eq: $$4 - i - 4w = 3i \Rightarrow -4w = 3i - 4 + i = 4i - 4 \Rightarrow w = 1 - i$$ 2. **Solve the system (ii):** $$z + w = 3i$$ $$2z + 3w = 2$$ From first: $z = 3i - w$ Substitute into second: $$2(3i - w) + 3w = 2 \Rightarrow 6i - 2w + 3w = 2 \Rightarrow w = 2 - 6i$$ Then: $$z = 3i - (2 - 6i) = -2 + 9i$$ 3. **Solve the system (iii):** $$3z + (2+i)w = 11 - i$$ $$(2 - i)z - w = -1 + i$$ From second: $$w = (2 - i)z - (-1 + i) = (2 - i)z + 1 - i$$ Substitute into first: $$3z + (2+i)((2 - i)z + 1 - i) = 11 - i$$ Calculate $(2+i)(2 - i) = 4 + 1 = 5$ $$3z + 5z + (2+i)(1 - i) = 11 - i$$ Calculate $(2+i)(1 - i) = 2 - 2i + i - i^2= 2 - i + 1= 3 - i$ So: $$8z + 3 - i = 11 - i \Rightarrow 8z = 8 \Rightarrow z = 1$$ Plug $z=1$ into $w$: $$w = (2 - i)(1) + 1 - i = 2 - i + 1 - i = 3 - 2i$$ 4. **Polynomials evaluations:** (i) $$P(z) = z^2 + 6z + 20$$ (ii) $$P(z) = 3z^2 + 7$$ (iii) $$P(z) = z^3 - 2z^2 + z - 2$$ 5. **Given roots:** $$z_1 = -1 + i, \ z_2 = -1 - i$$ Check for polynomial $$z^2 + 2z + 2 = 0$$: $$(z_1)^2 + 2z_1 + 2 = (-1 + i)^2 + 2(-1 + i) + 2 = (1 - 2i - 1) - 2 + 2i + 2 = 0$$ Similarly for $z_2$, validates roots. 6. **Solve quadratic:** $$z^2 - 2z + 5 = 0$$ Discriminant: $$D = (-2)^2 - 4(1)(5) = 4 - 20 = -16 < 0$$ Roots: $$z = \frac{2 \pm \sqrt{-16}}{2} = 1 \pm 2i$$ 7. **Solve quadratics with complex coefficients:** (i) $$z^2 + z + 3=0$$ Discriminant: $$1 - 12 = -11$$ Roots: $$z = \frac{-1 \pm \sqrt{-11}}{2} = \frac{-1}{2} \pm \frac{i\sqrt{11}}{2}$$ (ii) Rearrange $$z^2 - 1 = z \Rightarrow z^2 - z -1 =0$$ Discriminant: $$1 + 4 = 5$$ Roots: $$z = \frac{1 \pm \sqrt{5}}{2}$$ (iii) $$z^2 - 2z + i=0$$ Discriminant: $$4 - 4i$$ Using complex square root formula: $$\sqrt{4 - 4i} = 2 - i$$ (approx.) Roots: $$z = \frac{2 \pm (2 - i)}{2} = 2 - \frac{i}{2} \text{ or } \frac{i}{2}$$ (iv) $$z^2 + 4 = 0 \Rightarrow z^2 = -4$$ Roots: $$z = \pm 2i$$ 8. **Solve higher-degree equations:** (i) $$z^4 + z^2 + 1 = 0$$ Let $u = z^2$, $$u^2 + u + 1 = 0$$ Discriminant: $$1 - 4 = -3$$ $$u = \frac{-1 \pm i\sqrt{3}}{2}$$ Find $z$ such that $z^2 = u$ (complex roots). (ii) $$z^3 = -8$$ Cube root: $$z = 2 (\cos 180^\circ + i \sin 180^\circ), \text{ principal root } = -2$$ Other roots by De Moivre's theorem. (iii) $$(z - 1)^3 = -1$$ Let $w = z-1$, so $$w^3 = -1$$ Cube roots of $-1$ are: $$w = -1, \frac{1}{2} + i \frac{\sqrt{3}}{2}, \frac{1}{2} - i \frac{\sqrt{3}}{2}$$ Hence, $$z = 1 + w$$ (iv) $$z^3 = 1$$ Cube roots of unity are: $$1, \; \frac{-1 + i \sqrt{3}}{2}, \; \frac{-1 - i \sqrt{3}}{2}$$ Final answers are the above detailed roots and solutions for each problem set.