ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 677

\[1)\ \sqrt[4]{4}:\]

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

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

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

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

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

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

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

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

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

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

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

\[2)\ \sqrt[3]{i}:\]

\[z^{3} = i = 0 + i \bullet 1 =\]

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

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

\[z_{1} = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2}i;\]

\[z_{2} = \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} = - \frac{\sqrt{3}}{2} + \frac{1}{2}i;\]

\[z_{3} = \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2} = 0 - i = - i.\]

\[3)\ \sqrt[6]{1}:\]

\[z^{6} = 1 = 1 + i \bullet 0 =\]

\[= \cos{2\pi n} + i\sin{2\pi n};\]

\[z = \cos\frac{\text{πn}}{3} + i\sin\frac{\text{πn}}{3};\]

\[z_{1} = \cos 0 + i\sin 0 = 1 + i \bullet 0 = 1;\]

\[z_{2} = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i;\]

\[z_{3} = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} =\]

\[= - \frac{1}{2} + \frac{\sqrt{3}}{2}i;\]

\[z_{4} = \cos\pi + i\sin\pi =\]

\[= - 1 + i \bullet 0 = - 1;\]

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

\[= - \frac{1}{2} - \frac{\sqrt{3}}{2}i;\]

\[z_{6} = \cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3} = \frac{1}{2} - \frac{\sqrt{3}}{2}i.\]

\[4)\ \sqrt[4]{- 2 + 2i\sqrt{3}}:\]

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

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

\[z = \sqrt{2}\left( \cos\left( \frac{\pi}{6} + \frac{\text{πn}}{2} \right) + i\sin\left( \frac{\pi}{6} + \frac{\text{πn}}{2} \right) \right);\]

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

\[= \sqrt{2}\left( \frac{\sqrt{3}}{2} + \frac{1}{2}i \right) = \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i;\]

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

\[= \sqrt{2}\left( - \frac{1}{2} + \frac{\sqrt{3}}{2}i \right) = - \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2}i;\]

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

\[= \sqrt{2}\left( - \frac{\sqrt{3}}{2} - \frac{1}{2}i \right) = - \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}i;\]

\[z_{4} = \sqrt{2}\left( \cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3} \right) =\]

\[= \sqrt{2}\left( \frac{1}{2} - \frac{\sqrt{3}}{2}i \right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2}i.\]