ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 654

\[1)\ z^{4} + 81 = 0\]

\[z^{4} = - 81 = 81( - 1 + i \bullet 0) =\]

\[= 81\left( \cos(\pi + 2\pi n) + i\sin(\pi + 2\pi n) \right)\]

\[z = 3\left( \cos\frac{\pi + 2\pi n}{4} + i\sin\frac{\pi + 2\pi n}{4} \right)\]

\[z_{1} = 3\left( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \right) =\]

\[= 3\left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \right) = \frac{3\sqrt{2}}{2} + \frac{3\sqrt{2}}{2}i;\]

\[z_{2} = 3\left( \cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4} \right) =\]

\[= 3\left( - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \right) =\]

\[= - \frac{3\sqrt{2}}{2} + \frac{3\sqrt{2}}{2}i;\]

\[z_{3} = 3\left( \cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4} \right) =\]

\[= 3\left( - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \right) =\]

\[= - \frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2}i;\]

\[z_{4} = 3\left( \cos\frac{7\pi}{4} + i\sin\frac{7\pi}{4} \right) =\]

\[= 3\left( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \right) = \frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2}i.\]

\[2)\ 8z^{3} - 27 = 0\]

\[z^{3} = \frac{27}{8} = \frac{27}{8}(1 + i \bullet 0) =\]

\[= \frac{27}{8}\left( \cos{2\pi n} + i\sin{2\pi n} \right)\]

\[z = \frac{3}{2}\left( \cos\frac{2\pi n}{3} + i\sin\frac{2\pi n}{3} \right)\]

\[z_{1} = \frac{3}{2}\left( \cos 0 + i\sin 0 \right) =\]

\[= \frac{3}{2}(1 + i \bullet 0) = \frac{3}{2};\]

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

\[= \frac{3}{2}\left( - \frac{1}{2} + \frac{\sqrt{3}}{2} \right) = - \frac{3}{4} + \frac{3\sqrt{3}}{4}i;\]

\[z_{3} = \frac{3}{2}\left( \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} \right) =\]

\[= \frac{3}{2}\left( - \frac{1}{2} - \frac{\sqrt{3}}{2} \right) = - \frac{3}{4} - \frac{3\sqrt{3}}{4}i.\]

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

\[z^{4} = 0 + i \bullet 1 =\]

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

\[z = \cos\left( \frac{\pi}{8} + \frac{\text{πn}}{2} \right) + i\sin\left( \frac{\pi}{8} + \frac{\text{πn}}{2} \right)\]

\[z_{1} = \cos\frac{\pi}{8} + i\sin\frac{\pi}{8};\]

\[z_{2} = \cos\frac{5\pi}{8} + i\sin\frac{5\pi}{8};\]

\[z_{3} = \cos\frac{9\pi}{8} + i\sin\frac{9\pi}{8};\]

\[z_{4} = \cos\frac{13\pi}{8} + i\sin\frac{13\pi}{8}.\]

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

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

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

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

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

\[= \sqrt[3]{2}(0 + i \bullet 1) = i\sqrt[3]{2};\]

\[z_{2} = \sqrt[3]{2}\left( \cos\frac{7\pi}{6} + i\sin\frac{7\pi}{6} \right) =\]

\[= \sqrt[3]{2}\left( - \frac{\sqrt{3}}{2} - \frac{1}{2}i \right);\]

\[z_{3} = \sqrt[3]{2}\left( \cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6} \right) =\]

\[= \sqrt[3]{2}\left( \frac{\sqrt{3}}{2} - \frac{1}{2}i \right).\]

\[5)\ z^{3} = - 2 + 2i\]

\[z^{3} = \sqrt{8}\left( - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \right) =\]

\[= \sqrt{8}\left( \cos\left( \frac{3\pi}{4} + 2\pi n \right) + i\sin\left( \frac{3\pi}{4} + 2\pi n \right) \right)\]

\[z = \sqrt{2}\left( \cos\left( \frac{\pi}{4} + \frac{2\pi n}{3} \right) + i\sin\left( \frac{\pi}{4} + \frac{2\pi n}{3} \right) \right)\]

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

\[= \sqrt{2}\left( \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \right) = 1 + i;\]

\[z_{2} = \sqrt{2}\left( \cos\frac{11\pi}{12} + i\sin\frac{11\pi}{12} \right);\]

\[z_{3} = \sqrt{2}\left( \cos\frac{19\pi}{12} + i\sin\frac{19\pi}{12} \right).\]

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

\[z^{4} = \sqrt{2}\left( \frac{\sqrt{2}}{2} + i \bullet \frac{\sqrt{2}}{2} \right) =\]

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

\[z = \sqrt[8]{2}\left( \cos\left( \frac{\pi}{16} + \frac{\text{πn}}{2} \right) + i\sin\left( \frac{\pi}{16} + \frac{\text{πn}}{2} \right) \right)\]

\[z_{1} = \sqrt[8]{2}\left( \cos\frac{\pi}{16} + i\sin\frac{\pi}{16} \right);\]

\[z_{2} = \sqrt[8]{2}\left( \cos\frac{9\pi}{16} + i\sin\frac{9\pi}{16} \right);\]

\[z_{3} = \sqrt[8]{2}\left( \cos\frac{17\pi}{16} + i\sin\frac{17\pi}{16} \right);\]

\[z_{4} = \sqrt[8]{2}\left( \cos\frac{25\pi}{16} + i\sin\frac{25\pi}{16} \right).\]