ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 828

\[1)\ z^{2} - 25i = 0\]

\[z^{2} = 25i\]

\[(x + yi)^{2} = 25i\]

\[x^{2} - y^{2} + 2xyi = 25i\]

\[x^{2} - y^{2} = 0,\ \ \ 2xy = 25\]

\[x^{2} = y^{2},\ \ \ y = \frac{25}{2x}\]

\[x^{2} = \frac{625}{4x^{2}}\]

\[4x^{4} = 625\]

\[x^{4} = \frac{625}{4}\]

\[x = \pm \frac{5}{\sqrt{2}} = \pm \frac{5\sqrt{2}}{2};\]

\[y = \pm \frac{25}{2}\ :\frac{5\sqrt{2}}{2} = \pm \frac{5\sqrt{2}}{2}.\]

\[Ответ:\ \ z = \pm \left( \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}i \right).\]

\[2)\ z^{2} = - 8 + 6i\]

\[z^{2} = - 8 + 6i\]

\[(x + yi)^{2} = - 8 + 6i\]

\[x^{2} - y^{2} + 2xyi = - 8 + 6i\]

\[x^{2} - y^{2} = - 8\text{\ \ }\]

\[x^{2} - y^{2} + 8 = 0\text{\ \ \ }\]

\[2xy = 6\]

\[y = \frac{3}{x};\]

\[x^{2} - \frac{9}{x^{2}} + 8 = 0\]

\[x^{4} + 8x^{2} - 9 = 0\]

\[D = 64 + 36 = 100\]

\[x_{1}^{2} = \frac{- 8 - 10}{2} = - 9;\]

\[x_{2}^{2} = \frac{- 8 + 10}{2} = 1;\]

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

\[y = \frac{3}{\pm 1} = \pm 3.\]

\[Ответ:\ \ z = \pm (1 + 3i).\]

\[3)\ z^{3} + 8 = 0\]

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

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

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

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

\[x = - 2.\]

\[Ответ:\ - 2;\ 1 \pm i\sqrt{3}.\]

\[4)\ z^{4} - 1 = 0\]

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

\[z_{1}^{2} = - 1;\text{\ \ \ }z_{2}^{2} = 1;\]

\[z_{1} = \pm \sqrt{- 1} = \pm i;\]

\[z_{2} = \pm \sqrt{1} = \pm 1.\]

\[Ответ:\ \pm 1;\ \pm i.\]