1) Z₄ + Z₂ = 2) Z₃ - Z₁ + 10 = 3) Z₂ * Z₁ = 4) Z₁ * (Z₄ + Z₃) = 5) Z₁ / Z₃ = 6) Z₄ / Z₂ = Z₁ = 4 - 2i Z₂ = 3 + 3i Z₃ = 1 - 5i Z₄ = -2 + 6i

Решение:

1. \( Z_4 + Z_2 = (-2 + 6i) + (3 + 3i) = (-2+3) + (6+3)i = 1 + 9i \)
2. \( Z_3 - Z_1 + 10 = (1 - 5i) - (4 - 2i) + 10 = 1 - 5i - 4 + 2i + 10 = (1-4+10) + (-5+2)i = 7 - 3i \)
3. \( \overline{Z_2} \cdot Z_1 = (3 - 3i) \cdot (4 - 2i) = (3 \cdot 4 + (-3) \cdot (-2)) + (3 \cdot (-2) + (-3) \cdot 4)i = (12 + 6) + (-6 - 12)i = 18 - 18i \)
4. \( Z_1 \cdot (Z_4 + Z_3) = (4 - 2i) \cdot ((-2 + 6i) + (1 - 5i)) = (4 - 2i) \cdot ((-2+1) + (6-5)i) = (4 - 2i) \cdot (-1 + i) = (4 \cdot (-1) + (-2) \cdot 1) + (4 \cdot 1 + (-2) \cdot (-1))i = (-4 - 2) + (4 + 2)i = -6 + 6i \)
5. \( \frac{Z_1}{Z_3} = \frac{4 - 2i}{1 - 5i} = \frac{(4 - 2i)(1 + 5i)}{(1 - 5i)(1 + 5i)} = \frac{4 + 20i - 2i - 10i^2}{1^2 - (5i)^2} = \frac{4 + 18i + 10}{1 - 25i^2} = \frac{14 + 18i}{1 + 25} = \frac{14 + 18i}{26} = \frac{14}{26} + \frac{18}{26}i = \frac{7}{13} + \frac{9}{13}i \)
6. \( \frac{Z_4}{Z_2} = \frac{-2 + 6i}{3 + 3i} = \frac{(-2 + 6i)(3 - 3i)}{(3 + 3i)(3 - 3i)} = \frac{-6 + 6i + 18i - 18i^2}{3^2 - (3i)^2} = \frac{-6 + 24i + 18}{9 - 9i^2} = \frac{12 + 24i}{9 + 9} = \frac{12 + 24i}{18} = \frac{12}{18} + \frac{24}{18}i = \frac{2}{3} + \frac{4}{3}i \)

Ответ: 1) \( 1 + 9i \), 2) \( 7 - 3i \), 3) \( 18 - 18i \), 4) \( -6 + 6i \), 5) \( \frac{7}{13} + \frac{9}{13}i \), 6) \( \frac{2}{3} + \frac{4}{3}i \).