Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 647

\[1)\ z_{1} = 2 + \frac{1}{2}i;\ z_{2} = \overline{z_{1}} = 2 - \frac{1}{2}i:\]

\[\left( z - 2 - \frac{1}{2}i \right)\left( z - 2 + \frac{1}{2}i \right) = 0\]

\[(z - 2)^{2} - \frac{1}{4}i^{2} = 0\]

\[z^{2} - 4z + 4 + \frac{1}{4} = 0\]

\[z^{2} - 4z + \frac{17}{4} = 0\]

\[D = 16 - 17 = - 1\]

\[z = \frac{4 \pm \sqrt{- 1}}{2} = \frac{4 \pm i}{2} = 2 \pm \frac{1}{2}i.\]

\[2)\ z_{1} = \frac{1}{2} - \frac{1}{2}i;\ z_{2} = \overline{z_{1}} = \frac{1}{2} + \frac{1}{2}i:\]

\[\left( z - \frac{1}{2} + \frac{1}{2}i \right)\left( z - \frac{1}{2} - \frac{1}{2}i \right) = 0\]

\[\left( z - \frac{1}{2} \right)^{2} - \frac{1}{4}i^{2} = 0\]

\[z^{2} - z + \frac{1}{4} + \frac{1}{4} = 0\]

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

\[D = 1 - 2 = - 1\]

\[z = \frac{1 \pm \sqrt{- 1}}{2} = \frac{1 \pm i}{2} = \frac{1}{2} \pm \frac{1}{2}i.\]

\[3)\ z_{1} = \sqrt{2} - \sqrt{5}i;\text{\ \ \ }\]

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

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

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

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

\[z^{2} - 2\sqrt{2}z + 7 = 0\]

\[D = 8 - 28 = - 20\]

\[z = \frac{2\sqrt{2} \pm \sqrt{- 20}}{2} =\]

\[= \frac{2\sqrt{2} \pm 2\sqrt{5}i}{2} = \sqrt{2} \pm \sqrt{5}i.\]

\[4)\ z_{1} = - \sqrt{6} + \sqrt{3}i;\]

\[z_{2} = \overline{z_{1}} = - \sqrt{6} - \sqrt{3}i:\]

\[\left( z + \sqrt{6} - \sqrt{3}i \right)\left( z + \sqrt{6} + \sqrt{3}i \right) = 0\]

\[\left( z + \sqrt{6} \right)^{2} - 3i^{2} = 0\]

\[z^{2} + 2\sqrt{6}z + 6 + 3 = 0\]

\[z^{2} + 2\sqrt{6}z + 9 = 0\]

\[D = 24 - 36 = - 12\]

\[z = \frac{- 2\sqrt{6} \pm \sqrt{- 12}}{2} =\]

\[= \frac{- 2\sqrt{6} \pm 2\sqrt{3}i}{2} = - \sqrt{6} \pm \sqrt{3}i.\]