Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 606

\[1)\ z(3 - 2i) = 1 + 2i;\]

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

\[= \frac{3 + 2i + 6i + 4i^{2}}{9 - 4i^{2}} = \frac{3 + 8i - 4}{9 + 4} =\]

\[= \frac{- 1 + 8i}{13} = - \frac{1}{13} + \frac{8}{13}i.\]

\[2)\ z( - 3 + 2i) = 5 - 4i;\]

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

\[= \frac{10i + 15 - 8i^{2} - 12i}{4i^{2} - 9};\]

\[= \frac{15 - 2i + 8}{- 4 - 9} = \frac{23 - 2i}{- 13} =\]

\[= - \frac{23}{13} + \frac{2}{13}i.\]

\[3)\ z(1 - 3i) - 6 = 2i\]

\[z(1 - 3i) = 6 + 2i;\]

\[z = \frac{6 + 2i}{1 - 3i} = \frac{(6 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)} =\]

\[= \frac{6 + 18i + 2i + 6i^{2}}{1 - 9i^{2}} =\]

\[= \frac{6 + 20i - 6}{1 + 9} = \frac{20i}{10} = 2i.\]

\[4)\ z( - 1 + i) - 7i = - 1\]

\[z( - 1 + i) = - 1 + 7i;\]

\[z = \frac{- 1 + 7i}{- 1 + i} = \frac{(7i - 1)(i + 1)}{(i - 1)(i + 1)} =\]

\[= \frac{7i^{2} + 7i - i - 1}{i^{2} - 1} =\]

\[= \frac{- 7 + 6i - 1}{- 1 - 1} = \frac{- 8 + 6i}{- 2} =\]

\[= 4 - 3i.\]