1. **Problem (a): Solve the equation $2z - 3\overline{z} = \frac{-27 + 23i}{1 + i}$ where $z = x + yi$ and $\overline{z} = x - yi$.** 2. First, simplify the right-hand side by multiplying numerator and denominator by the conjugate of the denominator: $$\frac{-27 + 23i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(-27 + 23i)(1 - i)}{(1 + i)(1 - i)}$$ 3. Calculate denominator: $$(1 + i)(1 - i) = 1 - i + i - i^2 = 1 - (-1) = 2$$ 4. Calculate numerator: $$(-27)(1) + (-27)(-i) + 23i(1) - 23i(i) = -27 + 27i + 23i - 23i^2$$ Since $i^2 = -1$, this becomes: $$-27 + 27i + 23i + 23 = (-27 + 23) + (27i + 23i) = -4 + 50i$$ 5. So the right side is: $$\frac{-4 + 50i}{2} = -2 + 25i$$ 6. Write $z = x + yi$, $\overline{z} = x - yi$. Substitute into the left side: $$2z - 3\overline{z} = 2(x + yi) - 3(x - yi) = 2x + 2yi - 3x + 3yi = (2x - 3x) + (2y + 3y)i = -x + 5yi$$ 7. Equate real and imaginary parts: $$-x = -2 \implies x = 2$$ $$5y = 25 \implies y = 5$$ 8. **Answer (a):** $$z = 2 + 5i$$ --- 1. **Problem (b)(i): Use De Moivre's Theorem to find the three roots of $z^3 = -4 + 4\sqrt{3}i$.** 2. Express the complex number in polar form: Calculate modulus: $$r = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 16 \times 3} = \sqrt{16 + 48} = \sqrt{64} = 8$$ 3. Calculate argument $\theta$: $$\theta = \arctan\left(\frac{4\sqrt{3}}{-4}\right) = \arctan(-\sqrt{3})$$ Since real part is negative and imaginary part positive, $z$ is in the second quadrant, so: $$\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$ 4. So: $$z^3 = 8 \left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right)$$ 5. To find the cube roots, use De Moivre's Theorem: $$z_k = r^{1/3} \left( \cos \frac{\theta + 2k\pi}{3} + i \sin \frac{\theta + 2k\pi}{3} \right), \quad k=0,1,2$$ 6. Calculate $r^{1/3}$: $$8^{1/3} = 2$$ 7. Calculate roots: - For $k=0$: $$z_0 = 2 \left( \cos \frac{2\pi/3}{3} + i \sin \frac{2\pi/3}{3} \right) = 2 \left( \cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9} \right)$$ - For $k=1$: $$z_1 = 2 \left( \cos \frac{2\pi/3 + 2\pi}{3} + i \sin \frac{2\pi/3 + 2\pi}{3} \right) = 2 \left( \cos \frac{8\pi}{9} + i \sin \frac{8\pi}{9} \right)$$ - For $k=2$: $$z_2 = 2 \left( \cos \frac{2\pi/3 + 4\pi}{3} + i \sin \frac{2\pi/3 + 4\pi}{3} \right) = 2 \left( \cos \frac{14\pi}{9} + i \sin \frac{14\pi}{9} \right)$$ 8. **Answer (b)(i):** $$z_1 = 2 \left( \cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9} \right), \quad z_2 = 2 \left( \cos \frac{8\pi}{9} + i \sin \frac{8\pi}{9} \right), \quad z_3 = 2 \left( \cos \frac{14\pi}{9} + i \sin \frac{14\pi}{9} \right)$$