Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 731

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

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

\[= - 1 + \frac{9i + 4i}{6} + 1 = \frac{13}{6}i.\]

\[2)\ \left( - 5 + \sqrt{2}i \right)\left( - 6 - 3\sqrt{2}i \right) =\]

\[= 30 + 15\sqrt{2}i - 6\sqrt{2}i - 6i^{2} =\]

\[= 30 + 9\sqrt{2}i + 6 = 36 + 9\sqrt{2}i.\]

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

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

\[= - 2 - 2i.\]

\[4)\ (3 - 2i)(4 + i) + 10i =\]

\[= 12 + 3i - 8i - 2i^{2} + 10i =\]

\[= 12 + 5i + 2 = 14 + 5i.\]

\[5)\ \frac{(3 - i)(1 + 3i)}{2 - i} =\]

\[= \frac{\left( 3 + 9i - i - 3i^{2} \right)(2 + i)}{(2 - i)(2 + i)} =\]

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

\[= \frac{(6 + 8i)(2 + i)}{4 + 1} =\]

\[= \frac{12 + 6i + 16i + 8i^{2}}{5} =\]

\[= \frac{12 + 22i - 8}{5} = \frac{4}{5} + \frac{22}{5}i.\]

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

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

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

\[= \frac{8 + 4i - 12i - 6i^{2}}{16 - 4i^{2}} =\]

\[= \frac{8 - 8i + 6}{16 + 4} = \frac{7}{10} - \frac{2}{5}i.\]

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

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

\[= \frac{6 + 9i}{4 - 9i^{2}} + \frac{6 - 9i^{2}}{4 - 9i^{2}} = \frac{12}{4 + 9} = \frac{12}{13}.\]

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

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

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

\[= \frac{8 + 6i^{2}}{4 + 1} = \frac{8 - 6}{5} = \frac{2}{5}.\]