Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 624

\[1)\ z = 7:\]

\[tg\ \varphi = \frac{b}{a} = \frac{0}{7} = 0;\ \]

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

\[Ответ:\ \ 2\pi n.\]

\[2)\ z = - 4:\]

\[tg\ \varphi = \frac{b}{a} = \frac{0}{- 4} = 0;\ \]

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

\[Ответ:\ \ \pi + 2\pi n.\]

\[3)\ z = i:\]

\[a = 0;\]

\[\varphi = \frac{\pi}{2}.\]

\[Ответ:\ \ \frac{\pi}{2} + 2\pi n.\ \]

\[4)\ z = - 3i:\]

\[a = 0;\text{\ \ \ }\]

\[\varphi = \frac{\pi}{2}.\]

\[Ответ:\ - \frac{\pi}{2} + 2\pi n.\ \]

\[5)\ z = - 1 + i:\]

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

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

\[Ответ:\ \ \frac{3\pi}{4} + 2\pi n.\]

\[6)\ z = \sqrt{3} + i:\]

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

\[\varphi = arctg\frac{1}{\sqrt{3}} = \frac{\pi}{6}.\]

\[Ответ:\ \ \frac{\pi}{6} + 2\pi n.\]

\[7)\ z = 2 - 2i:\]

\[tg\ \varphi = \frac{b}{a} = \frac{- 2}{2} = - 1;\ \]

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

\[Ответ:\ - \frac{\pi}{4} + 2\pi n.\]

\[8)\ z = - 1 - \sqrt{3}i:\]

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

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

\[Ответ:\ \ \frac{4\pi}{3} + 2\pi n.\]