Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 679

\[1)\ z = (tg\ 2 - i)^{4}:\]

\[r = \sqrt{tg^{2}\ 2 + ( - 1)^{2}} =\]

\[= \sqrt{tg^{2}\ 2 + 1} = \sqrt{\frac{1}{\cos^{2}2}} = \frac{1}{\cos 2};\]

\[\cos\varphi = tg\ 2\ :\frac{1}{\cos 2} =\]

\[= \frac{\sin 2}{\cos 2} \bullet \cos 2 = \sin 2 =\]

\[= \cos\left( \frac{\pi}{2} - 2 \right);\]

\[\sin\varphi = - 1\ :\frac{1}{\cos 2} = - \cos 2 =\]

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

\[\sqrt[4]{z} = tg\ 2 - i =\]

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

\[z =\]

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

\[= \frac{1}{\cos^{4}2}\left( \cos(2\pi - 8) + i\sin(6\pi + 8) \right) =\]

\[= \frac{1}{\cos^{4}2}\left( \cos 8 + i\sin 8 \right).\]

\[2)\ z = \left( \sin\frac{6\pi}{5} + i\left( 1 + \cos\frac{6\pi}{5} \right) \right)^{5}:\]

\[r = \sqrt{\sin^{2}\frac{6\pi}{5} + \left( 1 + \cos\frac{6\pi}{5} \right)^{2}} =\]

\[= \sqrt{\sin^{2}\frac{6\pi}{5} + 1 + 2\cos\frac{6\pi}{5} + \cos^{2}\frac{6\pi}{5}} =\]

\[= \sqrt{2 + 2\cos\frac{6\pi}{5}} =\]

\[= \sqrt{4 \bullet \frac{1 + \cos\frac{6\pi}{5}}{2}} = \sqrt{4\cos^{2}\frac{3\pi}{5}} =\]

\[= - 2\cos\frac{3\pi}{5};\]

\[\cos\varphi = \frac{\sin\frac{6\pi}{5}}{- 2\cos\frac{3\pi}{5}} =\]

\[= \frac{2\sin\frac{3\pi}{5} \bullet \cos\frac{3\pi}{5}}{- 2\cos\frac{3\pi}{5}} = - \sin\frac{3\pi}{5} =\]

\[= \cos\frac{9\pi}{10};\]

\[\sin\varphi = \frac{1 + \cos\frac{6\pi}{5}}{- 2\cos\frac{3\pi}{5}} = - \frac{\cos^{2}\frac{3\pi}{5}}{\cos\frac{3\pi}{5}} =\]

\[= - \cos\frac{3\pi}{5} = \sin\frac{9\pi}{10};\]

\[\sqrt[5]{z} = \sin\frac{6\pi}{5} + i\left( 1 + \cos\frac{6\pi}{5} \right) =\]

\[= - 2\cos\frac{3\pi}{5}\left( \cos\frac{9\pi}{10} + i\sin\frac{9\pi}{10} \right);\]

\[z = - 32\cos^{5}\frac{3\pi}{5}\left( \cos\frac{9\pi}{2} + i\sin\frac{9\pi}{2} \right) =\]

\[= - 32\cos^{5}\frac{3\pi}{5}\left( \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} \right).\]