Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 596

\[1)\ z = 6 + 8i;\]

\[|z| = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} =\]

\[= \sqrt{100} = 10.\]

\[2)\ z = 8 - 6i;\]

\[|z| = \sqrt{8^{2} + ( - 6)^{2}} = \sqrt{64 + 36} =\]

\[= \sqrt{100} = 10.\]

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

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

\[= \sqrt{4} = 2.\]

\[4)\ z = \sqrt{2} - \sqrt{3}i;\]

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

\[= \sqrt{2 + 3} = \sqrt{5}.\]

\[5)\ z = 5i;\]

\[|z| = \sqrt{0^{2} + 5^{2}} = \sqrt{0 + 25} =\]

\[= \sqrt{25} = 5.\]

\[6)\ z = - 2i;\]

\[|z| = \sqrt{0^{2} + ( - 2)^{2}} = \sqrt{0 + 4} =\]

\[= \sqrt{4} = 2.\]

\[7)\ z = \frac{1}{4} - \frac{3}{4}i;\]

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

\[= \sqrt{\frac{1}{16} + \frac{9}{16}} = \sqrt{\frac{10}{16}} = \frac{\sqrt{10}}{4}.\]

\[8)\ z = - \frac{4}{7} - \frac{3}{7}i;\]

\[|z| = \sqrt{\left( - \frac{4}{7} \right)^{2} + \left( - \frac{3}{7} \right)^{2}} =\]

\[= \sqrt{\frac{16}{49} + \frac{9}{49}} = \sqrt{\frac{25}{49}} = \frac{5}{7}.\]