ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 793

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

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

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

\[= (1 + i)(0 + i \bullet 1) =\]

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

\[3)\ \frac{\cos\frac{7\pi}{10} + i\sin\frac{7\pi}{10}}{\cos\frac{\pi}{5} + i\sin\frac{\pi}{5}} =\]

\[= \cos\left( \frac{7\pi}{10} - \frac{\pi}{5} \right) + i\sin\left( \frac{7\pi}{10} - \frac{\pi}{5} \right) =\]

\[= \cos\frac{5\pi}{10} + i\sin\frac{5\pi}{10} =\]

\[= \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = 0 + i \bullet 1 = i;\]

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

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

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

\[= \frac{\sqrt{2} + \sqrt{2}i}{2} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i.\]