ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 635

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Год:2020-2021-2022-2023
Тип:учебник

Задание 635

\[1)\ \frac{i\left( \cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3} \right)}{\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}} =\]

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

\[= i\left( \cos\frac{9\pi}{6} + i\sin\frac{9\pi}{6} \right) =\]

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

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

\[2)\ \frac{1}{\cos\frac{4\pi}{3} - i\sin\frac{4\pi}{3}} =\]

\[= \frac{1 + i \bullet 0}{\cos\frac{4\pi}{3} + i\sin\left( - \frac{4\pi}{3} \right)} =\]

\[= \frac{\cos 0 + i\sin 0}{\cos\left( - \frac{4\pi}{3} \right) + \sin\left( - \frac{4\pi}{3} \right)} =\]

\[= \cos\left( 0 + \frac{4\pi}{3} \right) + i\sin\left( 0 + \frac{4\pi}{3} \right) =\]

\[= \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} = - \frac{1}{2} - \frac{\sqrt{3}}{2}i;\]

\[3)\ \frac{\left( \cos\frac{\pi}{3} - i\sin\frac{\pi}{3} \right)\left( 1 + \sqrt{3}i \right)^{7}}{i} =\]

\[= 128 \bullet \frac{\cos{2\pi} + i\sin{2\pi}}{i} =\]

\[= 128 \bullet \frac{1 + i \bullet 0}{i} = \frac{128}{i} =\]

\[= \frac{128i}{i^{2}} = - 128i;\]

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

\[= \frac{\cos 0 + i\sin 0}{1 - i} = \frac{1 + i \bullet 0}{1 - i} =\]

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

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

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

\[= \left( \frac{( - 1)^{4} + \sqrt{3}i \bullet ( - 1)^{2}}{4} \right)^{5} =\]

\[= \left( \frac{1}{2}\left( \frac{1}{2} + \frac{\sqrt{3}}{2}i \right) \right)^{5} =\]

\[= \left( \frac{1}{2} \right)^{5} \bullet \left( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \right)^{5} =\]

\[= \frac{1}{2^{5}}\left( \cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3} \right) =\]

\[= \frac{1}{32}\left( \frac{1}{2} - \frac{\sqrt{3}}{2}i \right) = \frac{1}{64} - \frac{\sqrt{3}}{64}i;\]

\[6)\ \frac{(2i)^{7}}{\left( - \sqrt{2} + i\sqrt{2} \right)^{6}} =\]

\[= \frac{\left( 2(0 + i \bullet 1) \right)^{7}}{\left( 2\left( - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \right) \right)^{6}} =\]

\[= \frac{2^{7}\left( \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} \right)^{7}}{2^{6}\left( \cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4} \right)^{6}} =\]

\[= 2\left( \cos( - \pi) + i\sin( - \pi) \right) =\]

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

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