1) Problem: Find the complex number $z$ satisfying $$4z - 3\overline{z} = \frac{2 + 4i}{1 - i}.$$ Step 1: Simplify the right side by multiplying numerator and denominator by the conjugate of the denominator: $$\frac{2 + 4i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(2 + 4i)(1 + i)}{(1 - i)(1 + i)}.$$ Step 2: Calculate numerator and denominator: Numerator: $(2)(1) + (2)(i) + (4i)(1) + (4i)(i) = 2 + 2i + 4i + 4i^2 = 2 + 6i + 4(-1) = 2 + 6i - 4 = -2 + 6i.$ Denominator: $1 - i^2 = 1 - (-1) = 2.$ Step 3: So the right side is $$\frac{-2 + 6i}{2} = -1 + 3i.$$ Step 4: Let $z = x + iy$, then $\overline{z} = x - iy$. Substitute into the equation: $$4(x + iy) - 3(x - iy) = -1 + 3i.$$ Step 5: Expand: $$4x + 4iy - 3x + 3iy = -1 + 3i,$$ which simplifies to $$(4x - 3x) + (4iy + 3iy) = -1 + 3i,$$ $$x + 7iy = -1 + 3i.$$ Step 6: Equate real and imaginary parts: Real: $x = -1$, Imaginary: $7y = 3 \Rightarrow y = \frac{3}{7}.$ Answer: $$z = -1 + \frac{3}{7}i.$$ 2) Problem: Express $$z = (2 - i)^2 + \frac{7 - 4i}{2 + i} - 8$$ in the form $x + iy$. Step 1: Calculate $(2 - i)^2$: $$(2 - i)^2 = 2^2 - 2 \times 2 \times i + i^2 = 4 - 4i + (-1) = 3 - 4i.$$ Step 2: Simplify $$\frac{7 - 4i}{2 + i}$$ by multiplying numerator and denominator by conjugate of denominator: $$\frac{7 - 4i}{2 + i} \times \frac{2 - i}{2 - i} = \frac{(7 - 4i)(2 - i)}{(2 + i)(2 - i)}.$$ Step 3: Calculate numerator: $$7 \times 2 - 7 \times i - 4i \times 2 + 4i \times i = 14 - 7i - 8i + 4i^2 = 14 - 15i + 4(-1) = 14 - 15i - 4 = 10 - 15i.$$ Denominator: $$2^2 - i^2 = 4 - (-1) = 5.$$ Step 4: So fraction is $$\frac{10 - 15i}{5} = 2 - 3i.$$ Step 5: Substitute back: $$z = (3 - 4i) + (2 - 3i) - 8 = (3 + 2 - 8) + (-4i - 3i) = -3 - 7i.$$ Answer: $$z = -3 - 7i.$$ 3) Problem: Solve $$z^2 = 21 - 20i$$ for complex $z$. Step 1: Let $z = a + bi$, then $$z^2 = (a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 - b^2 + 2ab i.$$ Step 2: Equate real and imaginary parts: $$a^2 - b^2 = 21,$$ $$2ab = -20 \Rightarrow ab = -10.$$ Step 3: From $ab = -10$, express $b = -\frac{10}{a}$. Substitute into first equation: $$a^2 - \left(-\frac{10}{a}\right)^2 = 21,$$ $$a^2 - \frac{100}{a^2} = 21.$$ Step 4: Multiply both sides by $a^2$: $$a^4 - 100 = 21a^2.$$ Step 5: Rearrange: $$a^4 - 21a^2 - 100 = 0.$$ Step 6: Let $x = a^2$, then $$x^2 - 21x - 100 = 0.$$ Step 7: Solve quadratic: $$x = \frac{21 \pm \sqrt{21^2 + 4 \times 100}}{2} = \frac{21 \pm \sqrt{441 + 400}}{2} = \frac{21 \pm \sqrt{841}}{2} = \frac{21 \pm 29}{2}.$$ Step 8: Two solutions: $$x_1 = \frac{21 + 29}{2} = 25, \quad x_2 = \frac{21 - 29}{2} = -4.$$ Since $a^2 = x$, $a^2$ must be non-negative, so $a^2 = 25$. Step 9: Then $a = \pm 5$. From $ab = -10$, If $a = 5$, then $b = -\frac{10}{5} = -2$. If $a = -5$, then $b = -\frac{10}{-5} = 2$. Step 10: Solutions: $$z = 5 - 2i \quad \text{or} \quad z = -5 + 2i.$$ 4) Problem: Plot complex numbers on the complex plane: (a) $z = 3 + 2i$ (b) $z = -8i$ (c) $z = 5 - 2i$ Answer: Points are at (3,2), (0,-8), and (5,-2) respectively on the Cartesian plane with real axis horizontal and imaginary axis vertical. 5) Problem: Find $x + iy$ given $$x(1 - i)^2 + y(2 + 3i)^2 = 6 - 10i.$$ Step 1: Calculate $(1 - i)^2$: $$(1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i.$$ Step 2: Calculate $(2 + 3i)^2$: $$(2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i.$$ Step 3: Substitute: $$x(-2i) + y(-5 + 12i) = 6 - 10i.$$ Step 4: Separate real and imaginary parts: Real: $-5y = 6,$ Imaginary: $-2x + 12y = -10.$ Step 5: Solve real part: $$y = -\frac{6}{5}.$$ Step 6: Substitute $y$ into imaginary part: $$-2x + 12 \left(-\frac{6}{5}\right) = -10,$$ $$-2x - \frac{72}{5} = -10,$$ $$-2x = -10 + \frac{72}{5} = -\frac{50}{5} + \frac{72}{5} = \frac{22}{5},$$ $$x = -\frac{11}{5}.$$ Answer: $$x + iy = -\frac{11}{5} - \frac{6}{5}i.$$ 6) Problem: Evaluate $$i \left( \frac{1}{2i} - \frac{1}{3i + 4} \right).$$ Step 1: Simplify each term: $$\frac{1}{2i} = \frac{1}{2i} \times \frac{-i}{-i} = \frac{-i}{-2i^2} = \frac{-i}{2}$$ since $i^2 = -1$. Step 2: Simplify $$\frac{1}{3i + 4}$$ by multiplying numerator and denominator by conjugate: $$\frac{1}{3i + 4} \times \frac{4 - 3i}{4 - 3i} = \frac{4 - 3i}{16 + 9} = \frac{4 - 3i}{25}.$$ Step 3: Substitute back: $$i \left( -\frac{i}{2} - \frac{4 - 3i}{25} \right) = i \left( -\frac{i}{2} - \frac{4}{25} + \frac{3i}{25} \right).$$ Step 4: Combine terms inside parentheses: $$= i \left( -\frac{i}{2} + \frac{3i}{25} - \frac{4}{25} \right) = i \left( i \left(-\frac{1}{2} + \frac{3}{25} \right) - \frac{4}{25} \right).$$ Step 5: Calculate coefficient of $i$: $$-\frac{1}{2} + \frac{3}{25} = -\frac{25}{50} + \frac{6}{50} = -\frac{19}{50}.$$ Step 6: So expression is $$i \left( i \left(-\frac{19}{50} \right) - \frac{4}{25} \right) = i \left( -\frac{19}{50} i - \frac{4}{25} \right).$$ Step 7: Distribute $i$: $$i \times -\frac{19}{50} i = -\frac{19}{50} i^2 = -\frac{19}{50} (-1) = \frac{19}{50},$$ $$i \times -\frac{4}{25} = -\frac{4}{25} i.$$ Step 8: Final result: $$\frac{19}{50} - \frac{4}{25} i.$$ 7) Problem: Solve $$\frac{1}{2 + i} - \frac{1}{3 + i} + 3i.$$ Step 1: Simplify each fraction by multiplying numerator and denominator by conjugate: $$\frac{1}{2 + i} \times \frac{2 - i}{2 - i} = \frac{2 - i}{4 + 1} = \frac{2 - i}{5} = \frac{2}{5} - \frac{1}{5} i,$$ $$\frac{1}{3 + i} \times \frac{3 - i}{3 - i} = \frac{3 - i}{9 + 1} = \frac{3 - i}{10} = \frac{3}{10} - \frac{1}{10} i.$$ Step 2: Substitute back: $$\left( \frac{2}{5} - \frac{1}{5} i \right) - \left( \frac{3}{10} - \frac{1}{10} i \right) + 3i = \left( \frac{2}{5} - \frac{3}{10} \right) + \left( -\frac{1}{5} i + \frac{1}{10} i + 3i \right).$$ Step 3: Calculate real part: $$\frac{2}{5} - \frac{3}{10} = \frac{4}{10} - \frac{3}{10} = \frac{1}{10}.$$ Step 4: Calculate imaginary part: $$-\frac{1}{5} + \frac{1}{10} + 3 = -\frac{2}{10} + \frac{1}{10} + 3 = -\frac{1}{10} + 3 = \frac{29}{10}.$$ Step 5: Final answer: $$\frac{1}{10} + \frac{29}{10} i.$$