Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 600

\[1)\ \frac{1 - i}{1 + i} = \frac{(1 - i)^{2}}{(1 + i)(1 - i)} =\]

\[= \frac{1 - 2i + i^{2}}{1 - i^{2}} = \frac{1 - 2i - 1}{1 + 1} =\]

\[= - \frac{2i}{2} = - i;\]

\[2)\ \frac{3 + 3i}{1 - 3i} = \frac{(3 + 3i)(1 + 3i)}{(1 - 3i)(1 + 3i)} =\]

\[= \frac{3 + 9i + 3i + 9i^{2}}{1 - 9i^{2}} =\]

\[= \frac{3 + 12i - 9}{1 + 9} = \frac{- 6 + 12i}{10} =\]

\[= - \frac{3}{5} + \frac{6}{5}i;\]

\[3)\ \frac{2i}{1 - i} = \frac{2i(1 + i)}{(1 - i)(1 + i)} =\]

\[= \frac{2i + 2i^{2}}{1 - i^{2}} = \frac{2i - 2}{1 + 1} = \frac{2i - 2}{2} =\]

\[= - 1 + i;\]

\[4)\ \frac{1 - i}{2i} = \frac{(1 - i) \bullet i}{2i \bullet i} = \frac{i - i^{2}}{2i^{2}} =\]

\[= \frac{i + 1}{- 2} = - \frac{1}{2} - \frac{1}{2}i;\]

\[5)\ \frac{2 + 5i}{- 1 + 6i} = \frac{(2 + 5i)(6i + 1)}{(6i - 1)(6i + 1)} =\]

\[= \frac{12i + 2 + 30i^{2} + 5i}{36i^{2} - 1} =\]

\[= \frac{2 + 17i - 30}{- 36 - 1} = \frac{- 28 + 17i}{- 37} =\]

\[= \frac{28}{37} - \frac{17}{37}i;\]

\[6)\ \frac{5}{- 1 - 2i} = \frac{5(1 - 2i)}{- (1 + 2i)(1 - 2i)} =\]

\[= - \frac{5 - 10i}{1 - 4i^{2}} = \frac{- 5 + 10i}{1 + 4} =\]

\[= \frac{- 5 + 10i}{5} = - 1 + 2i;\]

\[7)\ \frac{- 3 + 2i}{1 - 4i} = \frac{( - 3 + 2i)(1 + 4i)}{(1 - 4i)(1 + 4i)} =\]

\[= \frac{- 3 - 12i + 2i + 8i^{2}}{1 - 16i^{2}} =\]

\[= \frac{- 3 - 10i - 8}{1 + 16} = \frac{- 11 - 10i}{17} =\]

\[= - \frac{11}{17} - \frac{10}{17}i;\]

\[8)\ \frac{- 4 - 3i}{- 2 - 5i} = \frac{- (4 + 3i)(2 - 5i)}{- (2 + 5i)(2 - 5i)} =\]

\[= \frac{8 - 20i + 6i - 15i^{2}}{4 - 25i^{2}} =\]

\[= \frac{8 - 14i + 15}{4 + 25} = \frac{23 - 14i}{29} =\]

\[= \frac{23}{29} - \frac{14}{29}i.\]