Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 601

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

\[= \frac{\left( 8 - 4i - 6i + 3i^{2} \right)(1 - i)}{(1 + i)(1 - i)} =\]

\[= \frac{(8 - 10i - 3)(1 - i)}{1 - i^{2}} =\]

\[= \frac{(5 - 10i)(1 - i)}{1 + 1} =\]

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

\[= \frac{5 - 15i - 10}{2} = - \frac{5}{2} - \frac{15}{2}i;\]

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

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

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

\[= \frac{(1 + 3i)(3 + i)}{9 + 1} =\]

\[= \frac{3 + i + 9i + 3i^{2}}{10} =\]

\[= \frac{3 + 10i - 3}{10} = \frac{10i}{10} = i;\]

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

\[= \frac{1 - i}{6 - 8i + 3i - 4i^{2}} =\]

\[= \frac{1 - i}{6 - 5i + 4} = \frac{1 - i}{10 - 5i} =\]

\[= \frac{(1 - i)(10 + 5i)}{(10 - 5i)(10 + 5i)} =\]

\[= \frac{10 + 5i - 10i - 5i^{2}}{100 - 25i^{2}} =\]

\[= \frac{10 - 5i + 5}{100 + 25} = \frac{15 - 5i}{125} =\]

\[= \frac{3}{25} - \frac{1}{25}i;\]

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

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

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

\[= \frac{12 - 16i - 6i + 8i^{2}}{36 - 64i^{2}} =\]

\[= \frac{12 - 22i - 8}{36 + 64} = \frac{4 - 22i}{100} =\]

\[= \frac{1}{25} - \frac{11}{50}i;\]

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

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

\[= \frac{10 + 29i - 10}{4 + 25} + \frac{12 - 25i - 12}{16 + 9} =\]

\[= \frac{29i}{29} + \frac{- 25i}{25} = i - i = 0;\]

\[6)\ \frac{2 - 3i}{1 + 4i} - \frac{2 + 3i}{1 - 4i} =\]

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

\[= \frac{(2 - 11i - 12) - (2 + 11i - 12)}{1 + 16} =\]

\[= \frac{- 22i}{1 + 16} = - \frac{22}{17}i.\]