1. **Problem:** Given complex numbers $z_1 = 1 + i$, $z_2 = 3 - 2i$, and $z_3 = -2 + 3i$, calculate the following: 2. **Formula and rules:** - Multiplication: $(a+bi)(c+di) = (ac - bd) + (ad + bc)i$ - Conjugate: If $z = a + bi$, then $z^* = a - bi$ - Division: $\frac{z_1}{z_2} = \frac{z_1 z_2^*}{|z_2|^2}$ - Square: $z^2 = (a+bi)^2 = a^2 - b^2 + 2abi$ 3. **Calculations:** **a.** $z_1 \times z_2 + 2z_3^*$ - $z_1 z_2 = (1+i)(3-2i) = 3 - 2i + 3i - 2i^2 = 3 + i + 2 = 5 + i$ - $z_3^* = -2 - 3i$ - $2z_3^* = 2(-2 - 3i) = -4 - 6i$ - Sum: $(5 + i) + (-4 - 6i) = 1 - 5i$ **b.** $\frac{z_1}{z_3} - \frac{z_2}{5}$ - $|z_3|^2 = (-2)^2 + 3^2 = 4 + 9 = 13$ - $z_3^* = -2 - 3i$ - $\frac{z_1}{z_3} = \frac{(1+i)(-2 - 3i)}{13} = \frac{-2 - 3i - 2i - 3i^2}{13} = \frac{-2 - 5i + 3}{13} = \frac{1 - 5i}{13} = \frac{1}{13} - \frac{5}{13}i$ - $\frac{z_2}{5} = \frac{3 - 2i}{5} = \frac{3}{5} - \frac{2}{5}i$ - Difference: $\left(\frac{1}{13} - \frac{5}{13}i\right) - \left(\frac{3}{5} - \frac{2}{5}i\right) = \left(\frac{1}{13} - \frac{3}{5}\right) + \left(-\frac{5}{13} + \frac{2}{5}\right)i = -\frac{34}{65} + \frac{9}{65}i$ **c.** $z_1^2 - 3 z_2 z_3$ - $z_1^2 = (1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$ - $z_2 z_3 = (3 - 2i)(-2 + 3i) = -6 + 9i + 4i - 6i^2 = -6 + 13i + 6 = 0 + 13i$ - $3 z_2 z_3 = 3(0 + 13i) = 0 + 39i$ - Difference: $2i - 39i = -37i$ **d.** $\frac{z_1 z_3}{z_2^*}$ - $z_1 z_3 = (1+i)(-2 + 3i) = -2 + 3i - 2i + 3i^2 = -2 + i - 3 = -5 + i$ - $z_2^* = 3 + 2i$ - $|z_2|^2 = 3^2 + (-2)^2 = 9 + 4 = 13$ - $\frac{z_1 z_3}{z_2^*} = \frac{(-5 + i)(3 - 2i)}{13}$ - Numerator: $(-5)(3) + (-5)(-2i) + i(3) + i(-2i) = -15 + 10i + 3i - 2i^2 = -15 + 13i + 2 = -13 + 13i$ - Division: $\frac{-13 + 13i}{13} = -1 + i$ **e.** $\frac{2 z_1 - 4 z_2}{z_3 z_2^*}$ - $2 z_1 = 2(1 + i) = 2 + 2i$ - $4 z_2 = 4(3 - 2i) = 12 - 8i$ - Numerator: $(2 + 2i) - (12 - 8i) = 2 + 2i - 12 + 8i = -10 + 10i$ - $z_3 z_2^* = (-2 + 3i)(3 + 2i) = -6 - 4i + 9i + 6i^2 = -6 + 5i - 6 = -12 + 5i$ - Multiply numerator and denominator by conjugate of denominator: $$\frac{-10 + 10i}{-12 + 5i} \times \frac{-12 - 5i}{-12 - 5i} = \frac{(-10 + 10i)(-12 - 5i)}{(-12)^2 + 5^2} = \frac{120 + 50i - 120i - 50i^2}{144 + 25} = \frac{120 - 70i + 50}{169} = \frac{170 - 70i}{169} = \frac{170}{169} - \frac{70}{169}i$$ 4. **Final answers:** - a) $1 - 5i$ - b) $-\frac{34}{65} + \frac{9}{65}i$ - c) $-37i$ - d) $-1 + i$ - e) $\frac{170}{169} - \frac{70}{169}i$