ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 621

\[1)\ \left\{ \begin{matrix} |z + 1| = |z + 2|\text{\ \ \ \ \ \ \ \ \ } \\ |3z + 9| = |5z + 10i| \\ \end{matrix} \right.\ \]

\[|z + 1| = |z + 2|\ \]

\[(x + 1)^{2} + y^{2} = (x + 2)^{2} + y^{2}\]

\[x^{2} + 2x + 1 = x^{2} + 4x + 4\]

\[2x = - 3\]

\[x = - \frac{3}{2}.\]

\[|3z + 9| = |5z + 10i|\]

\[(3x + 9)^{2} + (3y)^{2} =\]

\[= (5x)^{2} + (5y + 10)^{2}\]

\[9x^{2} + 54x + 81 + 9y^{2} =\]

\[= 25x^{2} + 25y^{2} + 100y + 100\]

\[16y^{2} + 100y + 16x^{2} - 54x + 19 = 0\]

\[16y^{2} + 100y + 16 \bullet \frac{9}{4} + 54 \bullet \frac{3}{2} + 19 = 0\]

\[16y^{2} + 100y + 36 + 81 + 19 = 0\]

\[16y^{2} + 100y + 136 = 0\]

\[4y^{2} + 25y + 34 = 0\]

\[D = 625 - 544 = 81\]

\[y_{1} = \frac{- 25 - 9}{2 \bullet 4} = - \frac{17}{4};\text{\ \ }\]

\[y_{2} = \frac{- 25 + 9}{2 \bullet 4} = - 2.\]

\[Ответ:\ \ z_{1} = - \frac{3}{2} - \frac{17}{4}i;\ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }z_{2} = - \frac{3}{2} - 2i.\]

\[2)\ \left\{ \begin{matrix} (1 - i)\overline{z} = (1 + i)z \\ \left| z^{2} + 51i \right| = 1\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(1 - i)\overline{z} = (1 + i)z\]

\[(1 - i)(x - yi) = (1 + i)(x + yi)\]

\[x - yi - xi + yi^{2} = x + yi + xi + yi^{2}\]

\[2yi = - 2xi\]

\[y = - x.\]

\[\left| z^{2} + 51i \right| = 1\]

\[\left| (x + yi)^{2} + 51i \right| = 1\]

\[\left| (x - xi)^{2} + 51i \right| = 1\]

\[\left| x^{2} - 2x^{2}i + x^{2}i + 51i \right| = 1\]

\[\left| x^{2} - x^{2} + \left( 51 - 2x^{2} \right)i \right| = 1\]

\[\left| \left( 51 - 2x^{2} \right)i \right| = 1\]

\[51 - 2x^{2} = - 1\text{\ \ \ }\]

\[2x^{2} = 52\text{\ \ \ }\]

\[x^{2} = 26\text{\ \ }\]

\[x_{1} = \pm \sqrt{26}.\text{\ \ \ }\]

\[51 - 2x^{2} = 1\]

\[2x^{2} = 50\]

\[x^{2} = 25\]

\[x_{2} = \pm 5.\]

\[y_{1} = \mp \sqrt{26};\text{\ \ \ }y_{2} = \mp 5;\]

\[Ответ:\ \ \]

\[z_{1} = - \sqrt{26} + \sqrt{26}i;\ \]

\[z_{2} = \sqrt{26} - \sqrt{26}i;\]

\[z_{3} = - 5 + 5i;\ \]

\[z_{4} = 5 - 5i.\]