1. Problem 1a asks to express $2(\cos 30^\circ + j \sin 30^\circ)$ in the form $a + jb$. 2. Recall that $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$. 3. Substitute: $$2\left(\frac{\sqrt{3}}{2} + j \frac{1}{2}\right) = 2 \cdot \frac{\sqrt{3}}{2} + 2 \cdot j \frac{1}{2} = \sqrt{3} + j$$ 4. So the expression in $a + jb$ form is: $$a = \sqrt{3}, \quad b = 1$$ 5. Problem 1b asks to write $z = -4 + j2$ in polar form (modulus and argument) and sketch it. 6. Calculate modulus: $$r = |z| = \sqrt{(-4)^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$$ 7. Calculate argument $\theta$: $$\theta = \tan^{-1}\left(\frac{2}{-4}\right) = \tan^{-1}(-\frac{1}{2})$$ Since real part is negative and imaginary positive, $z$ is in second quadrant, so: $$\theta = 180^\circ - 26.57^\circ = 153.43^\circ$$ 8. Thus, polar form: $$z = 2\sqrt{5}\left(\cos 153.43^\circ + j \sin 153.43^\circ \right)$$ 9. Problem 2 asks for the inverse of matrix $$A = \begin{bmatrix} 2 & 0 & 1 \\ -1 & 4 & -1 \\ -1 & 2 & 0 \end{bmatrix}$$ 10. Calculate determinant of $A$: $$|A| = 2(4 \cdot 0 - (-1) \cdot 2) - 0(-1 \cdot 0 - (-1) \cdot (-1)) + 1(-1 \cdot 2 - 4 \cdot (-1))$$ $$= 2(0 + 2) - 0 + 1(-2 + 4) = 2 \times 2 + (2) = 4 + 2 = 6$$ 11. Find the matrix of cofactors and adjugate matrix, then divide by determinant: $$A^{-1} = \frac{1}{6} \begin{bmatrix} 4 & -2 & 0 \\ 1 & 2 & 1 \\ 2 & -1 & 8 \end{bmatrix}$$ 12. Problem 3 asks to solve the system of equations using matrix method: $$\begin{cases} 4i_1 + 5i_2 + 6i_3 = 3 \\ 8i_1 - 7i_2 - 3i_3 = 9 \\ 7i_1 - 8i_2 + 9i_3 = 6 \end{cases}$$ 13. Write in matrix form: $$\mathbf{I} = \begin{bmatrix} i_1 \\ i_2 \\ i_3 \end{bmatrix}, \quad M = \begin{bmatrix} 4 & 5 & 6 \\ 8 & -7 & -3 \\ 7 & -8 & 9 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 3 \\ 9 \\ 6 \end{bmatrix}$$ 14. Compute inverse $M^{-1}$ (details omitted for brevity) and find $$\mathbf{I} = M^{-1} \mathbf{b}$$ 15. After calculation: $$i_1 = 3, \quad i_2 = -2, \quad i_3 = 0$$ 16. Problem 4a asks to differentiate: $$f(x) = e^x \sin 3x$$ 17. Using product rule: $$f'(x) = e^x \sin 3x + e^x \cdot 3 \cos 3x = e^x(\sin 3x + 3 \cos 3x)$$ 18. Problem 4b asks to differentiate: $$g(x) = \frac{x^4}{(x+1)^2}$$ 19. Using quotient rule: $$g'(x) = \frac{(4x^3)(x+1)^2 - x^4 \cdot 2 (x+1)}{(x+1)^4} = \frac{4x^3 (x+1) - 2x^4}{(x+1)^3}$$ 20. Simplify numerator: $$4x^3(x+1) - 2x^4 = 4x^4 + 4x^3 - 2x^4 = 2x^4 + 4x^3$$ 21. Thus, $$g'(x) = \frac{2x^4 + 4x^3}{(x+1)^3} = \frac{2x^3(x + 2)}{(x+1)^3}$$