Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 631

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

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

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

\[tg\ \varphi = \frac{b}{a} = - \frac{\sqrt{2}}{2}\ :\left( - \frac{\sqrt{2}}{2} \right) = 1;\]

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

\[z = \cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}.\]

\[2) - 5\left( \cos{40{^\circ}} + i\sin{40{^\circ}} \right) =\]

\[= 5\left( - \cos{40{^\circ}} - i\sin{40{^\circ}} \right);\]

\[- \cos{40{^\circ}} < 0;\ - \sin{40{^\circ}} < 0;\]

\[\varphi = 180{^\circ} + 40{^\circ} = 220{^\circ}.\]

\[z = 5\left( \cos{220{^\circ}} + i\sin{220{^\circ}} \right).\]

\[3)\ 1 + \cos a + i\sin a;\ 0 \leq a \leq \frac{\pi}{2}:\]

\[|z| = \sqrt{\left( 1 + \cos a \right)^{2} + \sin^{2}a} =\]

\[= \sqrt{1 + 2\cos a + \cos^{2}a + \sin^{2}a} =\]

\[= \sqrt{2 + 2\cos a} = \sqrt{4 \bullet \frac{1 + \cos a}{2}} =\]

\[= \sqrt{4\cos^{2}\frac{a}{2}} = 2\cos\frac{a}{2};\]

\[\sin\varphi = \frac{b}{|z|} = \frac{\sin a}{2\cos\frac{a}{2}} =\]

\[= \frac{2\sin\frac{a}{2} \bullet \cos\frac{a}{2}}{2\cos\frac{a}{2}} = \sin\frac{a}{2}.\]

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

\[4)\ 1 + \cos\frac{12\pi}{7} + i\sin\frac{12\pi}{7}\]

\[|z| = \sqrt{\left( 1 + \cos\frac{12\pi}{7} \right)^{2} + \sin^{2}\frac{12\pi}{7}} =\]

\[= \sqrt{1 + 2\cos\frac{12\pi}{7} + 1} =\]

\[= \sqrt{2\left( 1 + \cos\frac{12\pi}{7} \right)} =\]

\[= \sqrt{4\cos^{2}\frac{6\pi}{7}} = - 2\cos\frac{6\pi}{7};\]

\[\sin\varphi = \frac{b}{|z|} = \frac{\sin\frac{12\pi}{7}}{- 2\cos\frac{6\pi}{7}} =\]

\[= \frac{2\sin{\frac{6\pi}{7} \bullet \cos\frac{6\pi}{7}}}{- 2\cos\frac{6\pi}{7}} = - \sin\frac{6\pi}{7} =\]

\[= \sin\left( \pi + \frac{6\pi}{7} \right) = \sin\frac{13\pi}{7}.\]

\[z = - 2\cos\frac{6\pi}{7}\left( \cos\frac{13\pi}{7} + i\sin\frac{13\pi}{7} \right).\]