ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 625

\[1)\ z = 3;\]

\[|z| = \sqrt{3^{2}} = \sqrt{9} = 3;\]

\[tg\ \varphi = \frac{b}{a} = \frac{0}{3} = 0;\ \]

\[\varphi = arctg\ 0 = 0.\]

\[Ответ:\ \ z = 3\left( \cos 0 + i\sin 0 \right).\]

\[2)\ z = - 1:\]

\[|z| = \sqrt{( - 1)^{2}} = \sqrt{1} = 1;\]

\[tg\ \varphi = \frac{b}{a} = \frac{0}{- 1} = 0;\ \]

\[\varphi = \pi + arctg\ 0 = \pi.\]

\[Ответ:\ \ z = \cos\pi + i\sin\pi.\]

\[3)\ z = 3 + 3i:\]

\[|z| = \sqrt{3^{2} + 3^{2}} = \sqrt{18} = 3\sqrt{2};\]

\[tg\ \varphi = \frac{b}{a} = \frac{3}{3} = 1;\ \]

\[\varphi = arctg\ 1 = \frac{\pi}{4}.\]

\[Ответ:\ \ \]

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

\[4)\ z = - 2 + 2\sqrt{3}i:\]

\[|z| = \sqrt{\left( 2\sqrt{3} \right)^{2} + ( - 2)^{2}} =\]

\[= \sqrt{16} = 4;\]

\[tg\ \varphi = \frac{b}{a} = \frac{2\sqrt{3}}{- 2} = - \sqrt{3};\ \]

\[\varphi = \pi + arctg\ \left( - \sqrt{3} \right) = \frac{2\pi}{3}.\]

\[Ответ:\ \ \]

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

\[5)\ z = - 1 - \sqrt{3}i:\]

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

\[= \sqrt{4} = 2;\]

\[tg\ \varphi = \frac{b}{a} = \frac{- \sqrt{3}}{- 1} = \sqrt{3};\ \]

\[\varphi = \pi + arctg\ \sqrt{3} = \frac{4\pi}{3}.\]

\[Ответ:\ \ \]

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

\[6)\ z = 5 - 5i:\]

\[|z| = \sqrt{5^{2} + ( - 5)^{2}} =\]

\[= \sqrt{50} = 5\sqrt{2};\]

\[tg\ \varphi = \frac{b}{a} = \frac{- 5}{5} = - 1;\ \]

\[\varphi = arctg\ ( - 1) = - \frac{\pi}{4}.\]

\[Ответ:\ \ \]

\[z = 5\sqrt{2}\left( \cos\left( - \frac{\pi}{4} \right) + i\sin\left( - \frac{\pi}{4} \right) \right).\]

\[7)\ z = \sqrt{5}\left( \cos\frac{\pi}{3} - i\sin\frac{\pi}{3} \right):\]

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

\[\varphi = - \frac{\pi}{3}.\]

\[Ответ:\ \ \]

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

\[8)\ z = - \cos\frac{\pi}{7} - i\sin\frac{\pi}{7}:\]

\[- \cos\frac{\pi}{7} < 0,\ \ \ - \sin\frac{\pi}{7} < 0;\]

\[\varphi = \pi + \frac{\pi}{7} = \frac{8\pi}{7};\]

\[Ответ:\ \ z = \cos\frac{8\pi}{7} + i\sin\frac{8\pi}{7}.\]