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МатематикаРешебник по алгебре и начала математического анализа 11 класс Колягин Задание 637
Задание 637
\[1)\ z^{2} = 16i;\]
\[\left( r\left( \cos\varphi + i\sin\varphi \right) \right)^{2} =\]
\[= 16(0 + i \bullet 1);\]
\[r^{2}\left( \cos{2\varphi} + i\sin{2\varphi} \right) =\]
\[= 16\left( \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} \right);\]
\[r^{2} = 16,\text{\ \ }\]
\[r = \pm 4.\text{\ \ }\]
\[2\varphi = \frac{\pi}{2}\]
\[\varphi = \frac{\pi}{4}.\]
\[z = \pm 4\left( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \right) =\]
\[= \pm \left( 2\sqrt{2} + 2\sqrt{2}i \right);\]
\[Ответ:\ \pm \left( 2\sqrt{2} + 2\sqrt{2}i \right).\]
\[2)\ z^{2} = - 4i;\]
\[\left( r\left( \cos\varphi + i\sin\varphi \right) \right)^{2} =\]
\[= 4\left( 0 + i \bullet ( - 1) \right);\]
\[r^{2}\left( \cos{2\varphi} + i\sin{2\varphi} \right) =\]
\[= 4\left( \cos\left( - \frac{\pi}{2} \right) + i\sin\left( - \frac{\pi}{2} \right) \right);\]
\[r^{2} = 4\]
\[r = \pm 2.\text{\ \ \ }\]
\[2\varphi = - \frac{\pi}{2}\]
\[\varphi = - \frac{\pi}{4}.\]
\[z = \pm 2\left( \cos\left( - \frac{\pi}{4} \right) + i\sin\left( - \frac{\pi}{4} \right) \right) =\]
\[= \pm \left( \sqrt{2} - \sqrt{2}i \right).\]
\[Ответ:\ \pm \left( \sqrt{2} - \sqrt{2}i \right).\]
\[3)\ z^{2} = 2 - 2i\sqrt{3};\]
\[\left( r\left( \cos\varphi + i\sin\varphi \right) \right)^{2} =\]
\[= 4\left( \frac{1}{2} - \frac{\sqrt{3}}{2}i \right);\]
\[r^{2}\left( \cos{2\varphi} + i\sin{2\varphi} \right) =\]
\[= 4\left( \cos\left( - \frac{\pi}{3} \right) + i\sin\left( - \frac{\pi}{3} \right) \right);\]
\[r^{2} = 4\text{\ \ \ }\]
\[r = \pm 2.\]
\[2\varphi = - \frac{\pi}{3}\text{\ \ }\]
\[\varphi = - \frac{\pi}{6}.\]
\[z = \pm 2\left( \cos\left( - \frac{\pi}{6} \right) + i\sin\left( - \frac{\pi}{6} \right) \right) =\]
\[= \pm \left( \sqrt{3} - i \right).\]
\[Ответ:\ \pm \left( \sqrt{3} - i \right).\]
\[4)\ z^{2} = - 1 - \sqrt{3}i\]
\[\left( r\left( \cos\varphi + i\sin\varphi \right) \right)^{2} =\]
\[= 2\left( - \frac{1}{2} - \frac{\sqrt{3}}{2}i \right);\]
\[r^{2}\left( \cos{2\varphi} + i\sin{2\varphi} \right) =\]
\[= 2\left( \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} \right);\]
\[r^{2} = 2,\ \ \ 2\varphi = \frac{4\pi}{3};\]
\[r = \pm \sqrt{2},\ \ \ \varphi = \frac{2\pi}{3};\]
\[z = \pm \sqrt{2}\left( \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} \right) =\]
\[= \pm \left( - \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2}i \right).\]
\[Ответ:\ \pm \left( \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2}i \right).\]
