Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 629

\[1)\ \frac{5 - i}{2 - 3i} = \frac{(5 - i)(2 + 3i)}{(2 - 3i)(2 + 3i)} =\]

\[= \frac{10 + 15i - 2i - 3i^{2}}{4 - 9i^{2}} =\]

\[= \frac{10 + 13i + 3}{4 + 9} = \frac{13 + 13i}{13} =\]

\[= 1 + i;\]

\[|z| = \sqrt{1^{2} + 1^{2}} = \sqrt{2};\]

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

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

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

\[2)\ \frac{i}{(1 + i)^{2}} = \frac{i}{1 + 2i + i^{2}} =\]

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

\[|z| = \sqrt{\left( \frac{1}{2} \right)^{2} + 0^{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2};\]

\[tg\ \varphi = \frac{b}{a} = 0\ :\frac{1}{2} = 0;\]

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

\[z = \frac{1}{2}\left( \cos 0 + i\sin 0 \right).\]

\[3)\ \frac{1 + \sqrt{3}i}{i^{3}} = \frac{\frac{1}{i} + \sqrt{3}}{i^{2}} =\]

\[= - \sqrt{3} - \frac{i}{i^{2}} = - \sqrt{3} + i;\]

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

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

\[\varphi = arctg\left( - \frac{1}{\sqrt{3}} \right) = - \frac{\pi}{6}.\]

\[z = 2\left( \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} \right).\]

\[4)\ \frac{(1 + i)\left( 2 + i^{5} \right)}{3 - i^{13}} =\]

\[= \frac{(1 + i)(2 + i)}{3 - i} = \frac{2 + i + 2i + i^{2}}{3 - i} =\]

\[= \frac{2 + 3i - 1}{3 - i} = \frac{(1 + 3i)(3 + i)}{(3 - i)(3 + i)} =\]

\[= \frac{3 + i + 9i + 3i^{2}}{9 - i^{2}} =\]

\[= \frac{3 + 10i - 3}{9 + 1} = \frac{10i}{10} = i;\]

\[|z| = \sqrt{0^{2} + 1^{2}} = 1;\]

\[a = 0;\ \ \ \varphi = \frac{\pi}{2}.\]

\[\ z = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}.\]