ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 652

\[1)\ z^{2} = - 5 + 12i\]

\[(x + yi)^{2} = - 5 + 12i\]

\[x^{2} + 2xyi - y^{2} = - 5 + 12i\]

\[2xy = 12\text{\ \ \ }\]

\[y = \frac{6}{x}.\text{\ \ \ }\]

\[x^{2} - y^{2} = - 5\]

\[x^{2} - \frac{36}{x^{2}} = - 5\]

\[x^{4} + 5x^{2} - 36 = 0\]

\[D = 25 + 144 = 169\]

\[x_{1}^{2} = \frac{- 5 - 13}{2} = - 9;\text{\ \ }\]

\[x_{2}^{2} = \frac{- 5 + 13}{2} = 4;\]

\[x = \pm \sqrt{4} = \pm 2;\]

\[y = \frac{6}{\pm 2} = \pm 3.\]

\[Ответ:\ \pm (2 + 3i).\]

\[2)\ z^{2} = - 3 - 4i\]

\[(x + yi)^{2} = - 3 - 4i\]

\[x^{2} + 2xyi - y^{2} = - 3 - 4i\]

\[2xy = - 4\]

\[y = - \frac{2}{x}.\]

\[x^{2} - y^{2} = - 3\]

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

\[x^{4} + 3x^{2} - 4 = 0\]

\[D = 9 + 16 = 25\]

\[x_{1}^{2} = \frac{- 3 - 5}{2} = - 4;\text{\ \ }\]

\[x_{2}^{2} = \frac{- 3 + 5}{2} = 1;\]

\[x = \pm \sqrt{1} = \pm 1;\]

\[y = - \frac{2}{\pm 1} = \mp 2.\]

\[Ответ:\ \pm (1 - 2i).\]

\[3)\ z^{6} = 1\]

\[z^{3} = \pm \sqrt{1} = \pm 1\]

\[1)\ z^{3} = - 1\]

\[z^{3} + 1 = 0\]

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

\[D = 1 - 4 = - 3\]

\[z = \frac{1 \pm \sqrt{- 3}}{2} = \frac{1 \pm i\sqrt{3}}{2}.\]

\[2)\ z^{3} = 1\]

\[z^{3} - 1 = 0\]

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

\[D = 1 - 4 = - 3\]

\[z = \frac{- 1 \pm \sqrt{- 3}}{2} = \frac{- 1 \pm i\sqrt{3}}{2}.\]

\[Ответ:\ \pm 1;\ \frac{- 1 \pm i\sqrt{3}}{2};\ \frac{1 \pm i\sqrt{3}}{2}.\]

\[4)\ z^{6} = - 1\]

\[z^{3} = \pm \sqrt{- 1} = \pm i\]

\[1)\ z^{3} = - i\]

\[z^{3} + i = 0\]

\[z^{3} + iz^{2} - z - iz^{2} + z + i = 0\]

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

\[D = i^{2} + 4 \bullet 1 = - 1 + 4 = 3\]

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

\[2)\ z^{3} = i\]

\[z^{3} - i = 0\]

\[z^{3} - iz^{2} - z + iz^{2} + z - i = 0\]

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

\[D = i^{2} + 4 \bullet 1 = - 1 + 4 = 3\]

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

\[Ответ:\ \pm i;\ \frac{- \sqrt{3} \pm i}{2};\ \frac{\sqrt{3} \pm i}{2}.\]

\[5)\ z^{6} - 7z^{3} - 8 = 0\]

\[D = 49 + 32 = 81\]

\[z_{1}^{3} = \frac{7 - 9}{2} = - 1;\]

\[z_{2}^{3} = \frac{7 + 9}{2} = 8\]

\[1)\ z^{3} = - 1\]

\[z^{3} + 1 = 0\]

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

\[D = 1 - 4 = - 3\]

\[z = \frac{1 \pm \sqrt{- 3}}{2} = \frac{1 \pm i\sqrt{3}}{2}.\]

\[2)\ z^{3} = 8\]

\[z^{3} - 8 = 0\]

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

\[D = 4 - 16 = - 12\]

\[z = \frac{- 2 \pm \sqrt{- 12}}{2} = \frac{- 2 \pm 2\sqrt{3}i}{2} =\]

\[= - 1 \pm i\sqrt{3}.\]

\[Ответ:\ - 1;\ 2;\ \frac{1 \pm i\sqrt{3}}{2};\ - 1 \pm i\sqrt{3}.\]