Решебник по алгебре и начала математического анализа 11 класс Колягин Проверь себя VII. Сложный вариант

\[\mathbf{6}\text{.\ }\]

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

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

\[\mathbf{7}\text{.\ }\]

\[1)\ \]

\[\left( \cos{18{^\circ}} + i\sin{18{^\circ}} \right)\left( \cos{42{^\circ}} + i\sin{42{^\circ}} \right) =\]

\[= \cos(18{^\circ} + 42{^\circ}) + i\sin(18{^\circ} + 42{^\circ}) =\]

\[= \cos{60{^\circ}} + i\sin{60{^\circ}} = \frac{1}{2} + \frac{\sqrt{3}}{2}i;\]

\[2)\ \frac{9\left( \cos\frac{7\pi}{3} + i\sin\frac{7\pi}{3} \right)}{18\left( \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} \right)} =\]

\[= \frac{9}{18}\left( \cos\left( \frac{7\pi}{3} - \frac{4\pi}{3} \right) + i\sin\left( \frac{7\pi}{3} - \frac{4\pi}{3} \right) \right) =\]

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

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

\[3)\ \left( \sqrt{5}\left( \cos\frac{\pi}{8} + i\sin\frac{\pi}{8} \right) \right)^{4} =\]

\[= \left( \sqrt{5} \right)^{4} \bullet \left( \cos\frac{4\pi}{8} + i\sin\frac{4\pi}{8} \right) =\]

\[= 25\left( \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} \right) =\]

\[= 25(0 + i) = 25i.\]

\[\mathbf{8.}\]

\[1)\ z^{4} = 16 = 16(1 + 0i)\]

\[z^{4} = 16\left( \cos{2\pi} + i\sin{2\pi} \right)\]

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

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

\[= 2(0 + i) = 2i;\]

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

\[= 2( - 1 + 0i) = - 2;\]

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

\[= 2(0 - i) = - 2i;\]

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

\[= 2(1 + 0i) = 2.\]

\[Ответ:\ - 2i;\ 2i;\ - 2;\ 2.\]

\[2)\ z^{3} = 64 = 64(1 + 0i)\]

\[z^{3} = 64\left( \cos{2\pi} + i\sin{2\pi} \right)\]

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

\[z_{1} = 4\left( \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} \right) =\]

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

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

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

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

\[= 4(1 + 0i) = 4.\]

\[Ответ:\ \]

\[- 2 - 2\sqrt{3}i;\ - 2 + 2\sqrt{3}i;\ 4.\]