1. Solve the system: (i) $z - 4w = 3i$ and $2z + 3w = 11 - 5i$. Multiply first eq by 3: $3z - 12w = 9i$ Multiply second eq by 4: $8z + 12w = 44 - 20i$ Add: $11z = 44 - 11i \implies z = 4 - i$ Substitute in first eq: $4 - i - 4w = 3i \implies -4w = 3i - 4 + i = 4i - 4 \implies w = 1 - i$ 2. Solve system: (ii) $z + w = 3i$, $2z + 3w = 2$. From first: $z = 3i - w$ Substitute: $2(3i - w) + 3w = 2 \implies 6i - 2w + 3w = 2 \implies w = 2 - 6i$ Then $z = 3i - (2 - 6i) = -2 + 9i$ 3. Solve 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 $(2 + i)(2 - i) = 4 + 1 =5$ since $i^2 = -1$ So $3z + 5z + (2 + i)(1 - i) = 11 - i$ Calculate $(2 + i)(1 - i) = 2 - 2i + i - i^2 = 2 - i + 1 = 3 - i$ So: $8z + 3 - i = 11 - i$ implies $8z = 8$; $z = 1$ Then $w = (2 - i)*1 + 1 - i = 2 - i + 1 - i = 3 - 2i$ 4. Evaluate polynomials: (i) $P(z) = z^2 + 6z + 20$ (ii) $P(z) = 3z^2 + 7$ (iii) $P(z) = z^3 - 2z^2 + z - 2$ (Note: no specific z values given, so no numerical evaluation) 5. Verify $z_1 = -1 + i$ and $z_2 = -1 - i$ satisfy $z^2 + 2z + 2 = 0$. Calculate for $z_1$: $z_1^2 = (-1 + i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$ $2z_1 = 2(-1 + i) = -2 + 2i$ Sum: $-2i + (-2 + 2i) + 2 = (-2i + 2i) + (-2 + 2) = 0$ Similarly for $z_2$, the conjugate, also 0. 6. Check if $1 + 2i$ is a solution of $z^2 - 2z + 5 = 0$: $(1 + 2i)^2 - 2(1 + 2i) + 5 = (1 + 4i + 4i^2) - 2 - 4i + 5 = (1 + 4i - 4) - 2 - 4i + 5 = (-3 + 4i) - 2 - 4i + 5 = 0$ 7. Solve quadratic equations: (i) $z^2 + z + 3 = 0$ Discriminant: $1 - 12 = -11$ Roots: $z = \frac{-1 \pm i\sqrt{11}}{2}$ (ii) $z^2 - 1 = z$ or $z^2 - z - 1 = 0$ Discriminant: $1 + 4 = 5$ Roots: $z = \frac{1 \pm \sqrt{5}}{2}$ (iii) $z^2 - 2z + i = 0$ Discriminant: $4 - 4i$ Roots by formula: $z = \frac{2 \pm \sqrt{4 - 4i}}{2} = 1 \pm \frac{\sqrt{4 - 4i}}{2}$ (square root of complex can be found if needed) (iv) $z^2 + 4 = 0$ Roots: $z = \pm 2i$ 8. Solve higher-degree equations: (i) $z^4 + z^2 + 1 = 0$ Let $y = z^2$, then $y^2 + y + 1 = 0$ Discriminant: $1 - 4 = -3$ Roots $y = \frac{-1 \pm i\sqrt{3}}{2}$ Then $z = \pm \sqrt{y}$ complex root extraction (ii) $z^3 = -8$ Roots are cube roots of -8: Principal root: $z = 2 (\cos \pi + i \sin \pi) = -2$ Others: $z = 2 (\cos (\pi + 2k\pi/3) + i \sin (\pi + 2k\pi/3))$ for $k=1,2$ (iii) $(z - 1)^3 = -1$ Cube roots of -1 are $-1, \omega, \omega^2$ with $\omega = e^{2\pi i/3}$ So $z - 1 = -1$ or $z - 1 = \omega$ or $z - 1 = \omega^2$ Hence $z = 0, 1 + \omega, 1 + \omega^2$ (iv) $z^3 = 1$ Cube roots of unity: $1, \omega, \omega^2$ Final answers summarized in steps.