Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 673

\[1)\ z_{1} = \sqrt{3} - \sqrt{5}i;\text{\ \ \ }\]

\[z_{2} = \overline{z_{1}} = \sqrt{3} + \sqrt{5}i:\]

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

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

\[z^{2} - 2\sqrt{3}z + 3 + 5 = 0\]

\[z^{2} - 2\sqrt{3}z + 8 = 0.\]

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

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

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

\[= \frac{6 - 13i - 6}{4 + 9} = \frac{- 13i}{13} = - i;\text{\ \ \ }\]

\[z_{2} = \overline{z_{1}} = i:\]

\[(z + i)(z - i) = 0\]

\[z^{2} - i^{2} = 0\]

\[z^{2} + 1 = 0.\]