Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 651

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

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

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

\[= \frac{- 6 - 7i + 2}{4 + 1} = \frac{- 4 - 7i}{5} =\]

\[= - \frac{4}{5} - \frac{7}{5}i;\text{\ \ \ }\]

\[z_{2} = - \frac{4}{5} + \frac{7}{5}i:\]

\[\left( z + \frac{4}{5} + \frac{7}{5}i \right)\left( z + \frac{4}{5} - \frac{7}{5}i \right) = 0\]

\[\left( z + \frac{4}{5} \right)^{2} - \frac{49}{25}i^{2} = 0\]

\[z^{2} + \frac{8}{5}z + \frac{16}{25} + \frac{49}{25} = 0\]

\[z^{2} + \frac{8}{5}z + \frac{65}{25} = 0\]

\[5z^{2} + 8z + 13 = 0.\]

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

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

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

\[= \frac{- 4 - 3i - 1}{1 + 1} = \frac{- 5 - 3i}{2} =\]

\[= - \frac{5}{2} - \frac{3}{2}i;\text{\ \ \ }\]

\[z_{2} = - \frac{5}{2} + \frac{3}{2}i:\]

\[\left( z + \frac{5}{2} + \frac{3}{2}i \right)\left( z + \frac{5}{2} - \frac{3}{2}i \right) = 0\]

\[\left( z + \frac{5}{2} \right)^{2} - \frac{9}{4}i^{2} = 0\]

\[z^{2} + 5z + \frac{25}{4} + \frac{9}{4} = 0\]

\[z^{2} + 5z + \frac{34}{4} = 0\]

\[2z^{2} + 10z + 17 = 0.\]