22. Вычислить:

Пошаговое решение:

1. 1) \[ \frac{(4-3i)(2-i)}{1+i} = \frac{8 - 4i - 6i + 3i^2}{1+i} = \frac{8 - 10i - 3}{1+i} = \frac{5 - 10i}{1+i} = \frac{(5-10i)(1-i)}{(1+i)(1-i)} = \frac{5 - 5i - 10i + 10i^2}{1 - i^2} = \frac{5 - 15i - 10}{1+1} = \frac{-5 - 15i}{2} = -\frac{5}{2} - \frac{15}{2}i \]
2. 2) \[ \frac{(1+i)(2+i)}{3-i} = \frac{2 + i + 2i + i^2}{3-i} = \frac{2 + 3i - 1}{3-i} = \frac{1 + 3i}{3-i} = \frac{(1+3i)(3+i)}{(3-i)(3+i)} = \frac{3 + i + 9i + 3i^2}{9 - i^2} = \frac{3 + 10i - 3}{9+1} = \frac{10i}{10} = i \]
3. 3) \[ \frac{(2+i)(3-4i)}{1-i} = \frac{6 - 8i + 3i - 4i^2}{1-i} = \frac{6 - 5i + 4}{1-i} = \frac{10 - 5i}{1-i} = \frac{(10-5i)(1+i)}{(1-i)(1+i)} = \frac{10 + 10i - 5i - 5i^2}{1 - i^2} = \frac{10 + 5i + 5}{1+1} = \frac{15 + 5i}{2} = \frac{15}{2} + \frac{5}{2}i \]
4. 4) \[ \frac{(3-i)(1+3i)}{2-i} = \frac{3 + 9i - i - 3i^2}{2-i} = \frac{3 + 8i + 3}{2-i} = \frac{6 + 8i}{2-i} = \frac{(6+8i)(2+i)}{(2-i)(2+i)} = \frac{12 + 6i + 16i + 8i^2}{4 - i^2} = \frac{12 + 22i - 8}{4+1} = \frac{4 + 22i}{5} = \frac{4}{5} + \frac{22}{5}i \]
5. 5) \[ \frac{5+2i}{2-5i} + \frac{3-4i}{4+3i} = \frac{(5+2i)(2+5i)}{(2-5i)(2+5i)} + \frac{(3-4i)(4-3i)}{(4+3i)(4-3i)} = \frac{10 + 25i + 4i + 10i^2}{4 - 25i^2} + \frac{12 - 9i - 16i + 12i^2}{16 - 9i^2} = \frac{10 + 29i - 10}{4+25} + \frac{12 - 25i - 12}{16+9} = \frac{29i}{29} + \frac{-25i}{25} = i - i = 0 \]
6. 6) \[ \frac{2-3i}{1+4i} - \frac{2+3i}{1-4i} = \frac{(2-3i)(1-4i)}{(1+4i)(1-4i)} - \frac{(2+3i)(1+4i)}{(1-4i)(1+4i)} = \frac{2 - 8i - 3i + 12i^2}{1 - 16i^2} - \frac{2 + 8i + 3i + 12i^2}{1 - 16i^2} = \frac{2 - 11i - 12}{1+16} - \frac{2 + 11i - 12}{1+16} = \frac{-10 - 11i}{17} - \frac{-10 + 11i}{17} = \frac{-10 - 11i - (-10 + 11i)}{17} = \frac{-10 - 11i + 10 - 11i}{17} = \frac{-22i}{17} \]