Вычисли значения выражений с комплексными числами: a) Z1+Z2

Задание 1. а) $z_1 + z_2 = (4 - 2i) + (5 - 6i) = 4 - 2i + 5 - 6i = (4 + 5) + (-2i - 6i) = 9 - 8i$ б) $z_1 - z_2 = (4 - 2i) - (5 - 6i) = 4 - 2i - 5 + 6i = (4 - 5) + (-2i + 6i) = -1 + 4i$ в) $z_1 * z_2 = (4 - 2i) * (5 - 6i) = 4*5 + 4*(-6i) + (-2i)*5 + (-2i)*(-6i) = 20 - 24i - 10i + 12i^2 = 20 - 34i - 12 = 8 - 34i$ г) $\frac{z_1}{z_2} = \frac{4 - 2i}{5 - 6i} = \frac{(4 - 2i)(5 + 6i)}{(5 - 6i)(5 + 6i)} = \frac{4*5 + 4*6i - 2i*5 - 2i*6i}{5^2 + 6^2} = \frac{20 + 24i - 10i - 12i^2}{25 + 36} = \frac{20 + 14i + 12}{61} = \frac{32 + 14i}{61} = \frac{32}{61} + \frac{14}{61}i$ Задание 2. $(5 - 11i) + (6 - 3i)(8 - i) = 5 - 11i + 6*8 + 6*(-i) - 3i*8 - 3i*(-i) = 5 - 11i + 48 - 6i - 24i + 3i^2 = 5 - 11i + 48 - 6i - 24i - 3 = (5 + 48 - 3) + (-11i - 6i - 24i) = 50 - 41i$ Задание 3. $z^{-1} = \frac{1}{z} = \frac{1}{6 - i} = \frac{6 + i}{(6 - i)(6 + i)} = \frac{6 + i}{6^2 + 1^2} = \frac{6 + i}{36 + 1} = \frac{6 + i}{37} = \frac{6}{37} + \frac{1}{37}i$ **Ответ:** 1) a) $9 - 8i$ b) $-1 + 4i$ c) $8 - 34i$ d) $\frac{32}{61} + \frac{14}{61}i$ 2) $50 - 41i$ 3) $\frac{6}{37} + \frac{1}{37}i$