21. Найдите частное комплексных чисел:

Пошаговое решение:

1. 1) \[ \frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1 - 2i + i^2}{1 - i^2} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i \]
2. 2) \[ \frac{3+3i}{1-3i} = \frac{(3+3i)(1+3i)}{(1-3i)(1+3i)} = \frac{3 + 9i + 3i + 9i^2}{1 - 9i^2} = \frac{3 + 12i - 9}{1 + 9} = \frac{-6 + 12i}{10} = \frac{-3 + 6i}{5} \]
3. 3) \[ \frac{2i}{1-i} = \frac{2i(1+i)}{(1-i)(1+i)} = \frac{2i + 2i^2}{1 - i^2} = \frac{2i - 2}{1 + 1} = \frac{-2 + 2i}{2} = -1 + i \]
4. 4) \[ \frac{1-i}{2i} = \frac{(1-i)(-2i)}{(2i)(-2i)} = \frac{-2i + 2i^2}{-4i^2} = \frac{-2i - 2}{4} = \frac{-2 - 2i}{4} = \frac{-1 - i}{2} \]
5. 5) \[ \frac{2+5i}{-1+6i} = \frac{(2+5i)(-1-6i)}{(-1+6i)(-1-6i)} = \frac{-2 - 12i - 5i - 30i^2}{1 - 36i^2} = \frac{-2 - 17i + 30}{1 + 36} = \frac{28 - 17i}{37} \]
6. 6) \[ \frac{5}{-1-2i} = \frac{5(-1+2i)}{(-1-2i)(-1+2i)} = \frac{-5+10i}{1 - 4i^2} = \frac{-5+10i}{1+4} = \frac{-5+10i}{5} = -1+2i \]
7. 7) \[ \frac{-3+2i}{1-4i} = \frac{(-3+2i)(1+4i)}{(1-4i)(1+4i)} = \frac{-3 - 12i + 2i + 8i^2}{1 - 16i^2} = \frac{-3 - 10i - 8}{1 + 16} = \frac{-11 - 10i}{17} \]
8. 8) \[ \frac{-4-3i}{-2-5i} = \frac{(-4-3i)(-2+5i)}{(-2-5i)(-2+5i)} = \frac{8 - 20i + 6i - 15i^2}{4 - 25i^2} = \frac{8 - 14i + 15}{4 + 25} = \frac{23 - 14i}{29} \]