ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 636

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

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

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

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

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

\[2)\ z =\]

\[= - 3\left( \cos\left( \frac{\pi}{6} - \frac{\pi}{3} \right) + i\sin\left( \frac{\pi}{6} - \frac{\pi}{3} \right) \right) =\]

\[= - 3\left( \cos\left( - \frac{\pi}{6} \right) + i\sin\left( - \frac{\pi}{6} \right) \right) =\]

\[= 3\left( - \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} \right) =\]

\[= 3\left( \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} \right);\]

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

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

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

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

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

\[= \frac{4}{1 + 2i\sqrt{3} - 3} = \frac{4}{2i\sqrt{3} - 2} =\]

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

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

\[= \frac{2i\sqrt{3} + 2}{- 4} = - \frac{1}{2} - \frac{\sqrt{3}}{2}i =\]

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