Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 680

\[z^{10} - z^{5} - 992 = 0\]

\[D = 1 + 3968 = 3969\]

\[z_{1}^{5} = \frac{1 - 63}{2} = - 31;\]

\[z_{2}^{5} = \frac{1 + 63}{2} = 32;\]

\[1)\ z^{5} = - 31 = 31( - 1 + i \bullet 0) =\]

\[= 31\left( \cos(\pi + 2\pi n) + i\sin(\pi + 2\pi n) \right);\]

\[z = \sqrt[5]{31}\left( \cos\left( \frac{\pi}{5} + \frac{2\pi n}{5} \right) + i\sin\left( \frac{\pi}{5} + \frac{2\pi n}{5} \right) \right);\]

\[z_{1} = \sqrt[5]{31}\left( \cos\frac{3\pi}{5} + i\sin\frac{3\pi}{5} \right);\]

\[z_{2} = \sqrt[5]{31}\left( \cos\pi + i\sin\pi \right) = - \sqrt[5]{31};\]

\[z_{3} = \sqrt[5]{31}\left( \cos\frac{7\pi}{5} + i\sin\frac{7\pi}{5} \right).\]

\[2)\ z^{5} = 32 = 32(1 + i \bullet 0) =\]

\[= 32\left( \cos{2\pi n} + i\sin{2\pi n} \right);\]

\[z = 2\left( \cos\frac{2\pi n}{5} + i\sin\frac{2\pi n}{5} \right);\]

\[z_{1} = 2\left( \cos\frac{4\pi}{5} + i\sin\frac{4\pi}{5} \right);\]

\[z_{2} = 2\left( \cos\frac{6\pi}{5} + i\sin\frac{6\pi}{5} \right).\]

\[Ответ:\ \ \]

\[2\left( \cos\frac{4\pi}{5} + i\sin\frac{4\pi}{5} \right);\ \]

\[2\left( \cos\frac{6\pi}{5} + i\sin\frac{6\pi}{5} \right);\]

\[\sqrt[5]{31}\left( \cos\frac{3\pi}{5} + i\sin\frac{3\pi}{5} \right);\ \]

\[- \sqrt[5]{31};\ \sqrt[5]{31}\left( \cos\frac{7\pi}{5} + i\sin\frac{7\pi}{5} \right).\]