ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 678

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

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

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

\[2xy = - 16\ \]

\[y = - \frac{8}{x}.\]

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

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

\[x^{2} = \frac{64}{x^{2}}\]

\[x^{4} = 64\]

\[x = \pm \sqrt[4]{64} = \pm 2\sqrt{2};\]

\[y = - \frac{8}{\pm 2\sqrt{2}} = \mp \frac{4}{\sqrt{2}} = \mp 2\sqrt{2}.\]

\[z = \pm \left( 2\sqrt{2} - 2\sqrt{2}i \right).\]

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

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

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

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

\[y = \frac{3}{x}.\text{\ \ }\]

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

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

\[D = 64 + 36 = 100\]

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

\[x_{2}^{2} = \frac{8 + 10}{2} = 9.\]

\[x = \pm \sqrt{9} = \pm 3;\]

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

\[z = \pm (3 + i).\]

\[3)\ z^{3} = - 125\]

\[z^{3} + 125 = 0\]

\[(z + 5)\left( z^{2} - 5z + 25 \right) = 0\]

\[D = 25 - 100 = - 75\]

\[z_{1} = \frac{5 \pm \sqrt{- 75}}{2} = \frac{5 \pm 5\sqrt{3}i}{2} =\]

\[= \frac{5}{2} \pm \frac{5\sqrt{3}}{2}i;\]

\[z_{2} = - 5.\]

\[4)\ z^{4} = 16i\]

\[z^{4} = 16(0 + i) =\]

\[= 16\left( \cos\left( \frac{\pi}{2} + 2\pi n \right) + i\sin\left( \frac{\pi}{2} + 2\pi n \right) \right)\]

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

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

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

\[z_{3} = 2\left( \cos\frac{9\pi}{8} + i\sin\frac{9\pi}{8} \right);\]

\[z_{4} = 2\left( \cos\frac{13\pi}{8} + i\sin\frac{13\pi}{8} \right).\]

\[5)\ z^{3} - 1 = i\]

\[z^{3} = 1 + i = \sqrt{2}\left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \right) =\]

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

\[z = \sqrt[6]{2}\left( \cos\left( \frac{\pi}{12} + \frac{2\pi n}{3} \right) + i\sin\left( \frac{\pi}{12} + \frac{2\pi n}{3} \right) \right)\]

\[z_{1} = \sqrt[6]{2}\left( \cos\frac{\pi}{12} + i\sin\frac{\pi}{12} \right);\]

\[z_{2} = \sqrt[6]{2}\left( \cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4} \right);\]

\[z_{3} = \sqrt[6]{2}\left( \cos\frac{17\pi}{12} + i\sin\frac{17\pi}{12} \right).\]

\[6)\ z^{5} - 1 - i\sqrt{3} = 0\]

\[z^{5} = 1 + \sqrt{3}i = 2\left( \frac{1}{2} + \frac{\sqrt{3}}{2}i \right) =\]

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

\[z = \sqrt[5]{2}\left( \cos\left( \frac{\pi}{15} + \frac{2\pi n}{5} \right) + i\sin\left( \frac{\pi}{15} + \frac{2\pi n}{5} \right) \right)\]

\[z_{1} = \sqrt[5]{2}\left( \cos\frac{\pi}{15} + i\sin\frac{\pi}{15} \right);\]

\[z_{2} = \sqrt[5]{2}\left( \cos\frac{7\pi}{15} + i\sin\frac{7\pi}{15} \right);\]

\[z_{3} = \sqrt[5]{2}\left( \cos\frac{13\pi}{15} + i\sin\frac{13\pi}{15} \right);\]

\[z_{4} = \sqrt[5]{2}\left( \cos\frac{19\pi}{15} + i\sin\frac{19\pi}{15} \right);\]

\[z_{5} = \sqrt[5]{2}\left( \cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3} \right).\]