Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 594

\[1)\ z_{1} + z_{2} = z_{2} + z_{1}\]

\[\left( a_{1} + b_{1}i \right) + \left( a_{2} + b_{2}i \right) =\]

\[= \left( a_{2} + b_{2}i \right) + \left( a_{1} + b_{1}i \right)\]

\[\left( a_{1} + a_{2} \right) + \left( b_{1} + b_{2} \right)i =\]

\[= \left( a_{1} + a_{2} \right) + \left( b_{1} + b_{2} \right)i.\]

\[Что\ и\ требовалось\ доказать.\]

\[2)\ z_{1}z_{2} = z_{2}z_{1};\]

\[\left( a_{1} + b_{1}i \right)\left( a_{2} + b_{2}i \right) =\]

\[= \left( a_{2} + b_{2}i \right)\left( a_{1} + b_{1}i \right)\]

\[a_{1}a_{2} + a_{1}b_{2}i + a_{2}b_{1}i - b_{1}b_{2} =\]

\[= a_{1}a_{2} + a_{2}b_{1}i + a_{1}b_{2}i - b_{1}b_{2}\]

\[\left( a_{1}a_{2} - b_{1}b_{2} \right) + \left( a_{1}b_{2} + a_{2}b_{1} \right)i =\]

\[= \left( a_{1}a_{2} - b_{1}b_{2} \right) + \left( a_{1}b_{2} + a_{2}b_{1} \right)i.\]

\[Что\ и\ требовалось\ доказать.\]

\[3)\ \left( z_{1}z_{2} \right)z_{3} = z_{1}\left( z_{2}z_{3} \right);\]

\[\left( z_{1}z_{2} \right)z_{3} =\]

\[= \left( \left( a_{1} + b_{1}i \right)\left( a_{2} + b_{2}i \right) \right)\left( a_{3} + b_{3}i \right) =\]

\[z_{1}\left( z_{2}z_{3} \right) =\]

\[= \left( a_{1} + b_{1}i \right)\left( \left( a_{2} + b_{2}i \right)\left( a_{3} + b_{3}i \right) \right) =\]

\[Что\ и\ требовалось\ доказать.\]

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

\[\left( z_{1} + z_{2} \right) + z_{3} =\]

\[z_{1} + \left( z_{2} + z_{3} \right) =\]

\[Что\ и\ требовалось\ доказать.\]

\[5)\ z + 0 = z\]

\[(a + bi) + 0 = a + bi\]

\[a + bi = a + bi.\]

\[Что\ и\ требовалось\ доказать.\]

\[6)\ z \bullet 1 = z\]

\[(a + bi) \bullet 1 = a + bi\]

\[a + bi = a + bi.\]

\[Что\ и\ требовалось\ доказать.\]