ГДЗ по алгебре и начала математического анализа 11 класс Колягин Проверь себя VII

\[\mathbf{1.\ }\]

\[1)\ (3 + i) + (5 - 2i) = 8 - i;\]

\[2)\ (6 - i) - (2 + 3i) = 4 - 4i;\]

\[3)\ (7 + i)(10 - i) =\]

\[= 70 - 7i + 10i + 1 = 71 + 3i;\]

\[4)\ \frac{5 - 2i}{7 + 3i} = \frac{35 - 15i - 14i - 6}{49 + 9} =\]

\[= \frac{29 - 29i}{58} = \frac{1}{2} - \frac{1}{2}i.\]

\[\mathbf{2}\text{.\ }\]

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

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

\[= \frac{(4 + 10i - 25) + (4 - 10i - 25)}{4 + 25} =\]

\[= \frac{- 42}{29} = - 1\frac{13}{29}.\]

\[2)\ \left( \frac{1 + i^{11}}{2 - i^{7}} \right)^{2} = \left( \frac{1 - i}{2 + i} \right)^{2} =\]

\[= \frac{1 - 2i - 1}{4 + 4i - 1} = \frac{- 2i}{3 + 4i} =\]

\[= \frac{- 2i(3 - 4i)}{(3 + 4i)(3 - 4i)} = \frac{- 6i - 8}{9 + 16} =\]

\[= - \frac{8}{25} - \frac{6}{25}i.\]

\[\mathbf{3}\text{.\ }\]

\[1)\ z = - 1 + i\sqrt{3};\]

\[|r| = \sqrt{1 + 3} = \sqrt{4} = 2;\]

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

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

\[2)\ z = \sin\frac{\pi}{5} - i\cos\frac{\pi}{5} =\]

\[= \cos\left( \frac{\pi}{2} - \frac{\pi}{5} \right) - i\sin\left( \frac{\pi}{2} - \frac{\pi}{5} \right) =\]

\[= \cos\frac{3\pi}{10} - i\sin\frac{3\pi}{10} =\]

\[= \cos\left( - \frac{3\pi}{10} \right) + i\sin\left( - \frac{3\pi}{10} \right).\]

\[\mathbf{4}.\]

\[1)\ z^{2} + 5 = 0\]

\[z^{2} = - 5\]

\[z = \pm \sqrt{- 5} = \pm i\sqrt{5}.\]

\[Ответ:\ - i\sqrt{5};\ i\sqrt{5}.\]

\[2)\ z^{2} - 10z + 34 = 0;\]

\[D = 100 - 136 = - 36\]

\[z = \frac{10 \pm \sqrt{- 36}}{2} = \frac{10 \pm i\sqrt{6}}{2} =\]

\[= \frac{10 \pm 6i}{2} = 5 \pm 3i.\]

\[Ответ:\ \ 5 - 3i;\ 5 + 3i.\]

\[\mathbf{5}\text{.\ }\]

\[|z| = 3;\]

\[|x + yi| = 3;\]

\[\sqrt{x^{2} + y^{2}} = 3\]

\[x^{2} + y^{2} = 3^{2}.\]

\[Ответ:\ \ \]

\[окружность\ с\ центром\ (0;\ 0)\ \]

\[и\ R = 3.\]