Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 607

\[1)\ (3 - 2i)^{2} = 9 - 12i + 4i^{2} =\]

\[= 9 - 12i - 4 = 5 - 12i;\]

\[2)\ (1 + 2i)^{3} = 1 + 6i + 12i^{2} + 8i^{3} =\]

\[= 1 + 6i - 12 - 8i = - 11 - 2i;\]

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

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

\[= 4i^{2} = - 4;\]

\[4)\ (1 - i)^{6} = \left( 1 - 2i + i^{2} \right)^{3} =\]

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

\[= - 8i^{3} = 8i;\]

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

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

\[= 6i + 2i^{3} = 6i - 2i = 4i;\]

\[6)\ \left( \frac{1}{2} + \frac{i\sqrt{3}}{2} \right)^{2} + \left( \frac{1}{2} - \frac{i\sqrt{3}}{2} \right)^{2} =\]

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

\[= \frac{2}{4} + \frac{6i^{2}}{4} = \frac{2}{4} - \frac{6}{4} = - \frac{4}{4} = - 1.\]