Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 791

\[1)\ 2:\]

\[r = \sqrt{2^{2} + 0^{2}} = \sqrt{4} = 2;\]

\[\cos\varphi = \frac{a}{r} = \frac{2}{2} = 1 = \cos 0;\]

\[z = 2\left( \cos 0 + i\sin 0 \right).\]

\[2) - 3:\]

\[r = \sqrt{( - 3)^{2} + 0^{2}} = \sqrt{9} = 3;\]

\[\cos\varphi = \frac{a}{r} = \frac{- 3}{3} = - 1 = \cos\pi;\]

\[z = 3\left( \cos\pi + i\sin\pi \right).\]

\[3)\ 3i:\]

\[r = \sqrt{0^{2} + 3^{2}} = \sqrt{9} = 3;\]

\[\sin\varphi = \frac{b}{r} = \frac{3}{3} = 1 = \sin\frac{\pi}{2};\]

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

\[4) - 2i:\]

\[r = \sqrt{0^{2} + ( - 2)^{2}} = \sqrt{4} = 2;\]

\[\sin\varphi = \frac{b}{r} = \frac{- 2}{2} = - 1 = \sin\frac{3\pi}{2};\]

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

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

\[r = \sqrt{\left( \sqrt{3} \right)^{2} + ( - 1)^{2}} = \sqrt{4} = 2;\]

\[\cos\varphi = \frac{a}{r} = \frac{\sqrt{3}}{2} = \cos\frac{11\pi}{6};\]

\[\sin\varphi = \frac{b}{r} = \frac{- 1}{2} = - \frac{1}{2} = \sin\frac{11\pi}{6};\]

\[z = 2\left( \cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6} \right).\]

\[6)\ 2 - 2i:\]

\[r = \sqrt{2^{2} + ( - 2)^{2}} = \sqrt{8} = 2\sqrt{2};\]

\[\cos\varphi = \frac{a}{r} = \frac{2}{2\sqrt{2}} = \frac{\sqrt{2}}{2} = \cos\frac{7\pi}{4};\]

\[\sin\varphi = \frac{b}{r} = \frac{- 2}{2\sqrt{2}} = - \frac{\sqrt{2}}{2} = \sin\frac{7\pi}{4};\]

\[z = 2\sqrt{2}\left( \cos\frac{7\pi}{4} + i\sin\frac{7\pi}{4} \right).\]