ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 655

\[1)\ z^{2} = - 16 + 8i\]

\[(x + yi)^{2} = - 16 + 8i\]

\[x^{2} + 2xyi - y^{2} = - 16 + 8i\]

\[2xy = 8,\ \ \ x^{2} - y^{2} = - 16\]

\[y = \frac{4}{x},\ \ \ x^{2} - \frac{16}{x^{2}} = - 16\]

\[x^{4} + 16x^{2} - 16 = 0\]

\[D = 256 + 64 = 320\]

\[x = \frac{- 16 \pm \sqrt{320}}{2} = \frac{- 16 \pm 8\sqrt{5}}{2} =\]

\[= - 8 \pm 4\sqrt{5} = \pm \sqrt{- 8 + 4\sqrt{5}} =\]

\[= \pm 2\sqrt{\sqrt{5} - 2};\]

\[y = \frac{4}{\pm 2\sqrt{\sqrt{5} - 2}} = \pm \frac{2\sqrt{\sqrt{5} + 2}}{\sqrt{5 - 4}} =\]

\[= \pm 2\sqrt{\sqrt{5} + 2}.\]

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

\[2)\ 36z^{8} - 13z^{4} + 1 = 0\]

\[D = 169 - 144 = 25\]

\[z_{1}^{4} = \frac{13 - 5}{2 \bullet 36} = \frac{1}{9};\ \]

\[z_{2}^{4} = \frac{13 + 5}{2 \bullet 36} = \frac{1}{4};\]

\[z_{1}^{4} = \frac{1}{9}(1 + i \bullet 0) =\]

\[= \frac{1}{9}\left( \cos{2\pi n} + i\sin{2\pi n} \right);\]

\[z_{2}^{4} = \frac{1}{4}(1 + i \bullet 0) =\]

\[= \frac{1}{4}\left( \cos{2\pi n} + i\sin{2\pi n} \right);\]

\[z = \frac{1}{\sqrt{3}}\left( \cos\frac{\text{πn}}{2} + i\sin\frac{\text{πn}}{2} \right);\]

\[z = \frac{1}{\sqrt{2}}\left( \cos\frac{\text{πn}}{2} + i\sin\frac{\text{πn}}{2} \right);\]

\[z_{1} = \frac{1}{\sqrt{3}}\left( \cos 0 + i\sin 0 \right) =\]

\[= \frac{\sqrt{3}}{3}(1 + i \bullet 0) = \frac{\sqrt{3}}{3};\]

\[z_{2} = \frac{1}{\sqrt{2}}\left( \cos 0 + i\sin 0 \right) =\]

\[= \frac{\sqrt{2}}{2}(1 + i \bullet 0) = \frac{\sqrt{2}}{2};\]

\[z_{3} = \frac{1}{\sqrt{3}}\left( \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} \right) =\]

\[= \frac{\sqrt{3}}{3}(0 + i \bullet 1) = \frac{\sqrt{3}}{3}i;\]

\[z_{4} = \frac{1}{\sqrt{2}}\left( \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} \right) =\]

\[= \frac{\sqrt{2}}{2}(0 + i \bullet 1) = \frac{\sqrt{2}}{2}i;\]

\[z_{5} = \frac{1}{\sqrt{3}}\left( \cos\pi + i\sin\pi \right) =\]

\[= \frac{\sqrt{3}}{3}( - 1 + i \bullet 0) = - \frac{\sqrt{3}}{3};\]

\[z_{6} = \frac{1}{\sqrt{2}}\left( \cos\pi + i\sin\pi \right) =\]

\[= \frac{\sqrt{2}}{2}( - 1 + i \bullet 0) = - \frac{\sqrt{2}}{2};\]

\[z_{7} = \frac{1}{\sqrt{3}}\left( \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2} \right) =\]

\[= \frac{\sqrt{3}}{3}(0 - i) = - \frac{\sqrt{3}}{3}i;\]

\[z_{8} = \frac{1}{\sqrt{2}}\left( \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2} \right) =\]

\[= \frac{\sqrt{2}}{2}(0 - i) = - \frac{\sqrt{2}}{2}i.\]

\[3)\ z^{4} - 2z^{3} + 2z^{2} - 2z + 1 = 0\]

\[(z - 1)\left( z^{3} - z^{2} + z - 1 \right) = 0\]

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

\[z_{1} = 1;\text{\ \ \ }\]

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

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

\[Ответ:\ \ 1;\ \pm i.\]

\[4)\ z^{3} + \frac{1}{2}z^{2} + \frac{1}{2}z + 1 = 0\]

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

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

\[D = 1 - 16 = - 15\]

\[z = \frac{1 \pm \sqrt{- 15}}{2 \bullet 2} = \frac{1 \pm i\sqrt{15}}{4}.\]

\[Ответ:\ - 1;\ \frac{1 \pm i\sqrt{15}}{4}.\]