Предмет: Алгебра, автор: qewfaefaf

выполнить действия с комплексными числами: 3+7і и -2-5і, 4+3і и 2-7і.

Ответы

Автор ответа: mathkot

Ответ:

Действия с комплексными числами:

1) $z_{1} = 3 + 7i, z_{2} = -2 - 5i$

$z_{1} + z_{2} = 1 + 2i$

$z_{1} - z_{2}= 5 + 12i$

$z_{1}z_{2} = 29 - 29i$

$\displaystyle \frac{z_{1}}{z_{2}} = -\frac{41}{29} + i\frac{1}{29}$

2) $z_{1} = 4 + 3i, z_{2} = 2 - 7i$

$z_{1} + z_{2} = 6 - 4i$

$z_{1} - z_{2} = 2 + 10i$

$z_{1} z_{2} = 29 - 22i$

$\dfrac{z_{1}}{z_{2}} = -\dfrac{13}{53} + \dfrac{34}{53}i$

Примечание:

По определению: $i^{2} = -1$

Правила действия над комплексными числами в алгебраической форме $z_{1} = a_{1} + b_{1}i, z_{2} = a_{2} + b_{2}i$ :

Сложение:

$z_{1} + z_{2} = a_{1} + b_{1}i + a_{2} + b_{2}i = (a_{1} + a_{2}) + i(b_{1} + b_{2})$

Вычитание:

$z_{1} - z_{2} = a_{1} + b_{1}i - (a_{2} + b_{2}i) = (a_{1} - a_{2}) + i(b_{1} - b_{2})$

Умножение комплексного числа на действительное число:

$kz_{1} = k(a_{1} + b_{1}i) = ka_{1} + kb_{1}i, \ k \in \mathbb R$

Умножение комплексных чисел:

$z_{1}z_{2} = (a_{1} + b_{1}i)(a_{2} + b_{2}i) = a_{1}a_{2} + a_{1}b_{2}i + a_{2}b_{1}i + b_{2}b_{1}i^{2} =$

$=a_{1}a_{2} + a_{1}b_{2}i + a_{2}b_{1}i - b_{1}b_{2} = (a_{1}a_{2} - b_{1}b_{2} )+i(a_{1}b_{2} + a_{2}b_{1})$

Деление комплексных чисел $(z_{2} \neq 0)$ :

$\displaystyle \frac{z_{1}}{z_{2}} = \frac{a_{1} + b_{1}i}{a_{2} + b_{2}i} = \frac{(a_{1} + b_{1}i)(a_{2} - b_{2}i)}{(a_{2} + b_{2}i)(a_{2} - b_{2}i)} = \frac{a_{1}a_{2} - a_{1}b_{2}i + a_{2}b_{1}i - b_{1}b_{2}i^{2}}{a^{2}_{2} - i^{2}b_{2}^{2}} =$

$\displaystyle = \frac{a_{1}a_{2}+ b_{1}b_{2} + i(a_{2}b_{1} - a_{1}b_{2})}{a^{2}_{2} + b_{2}^{2}} = \frac{a_{1}a_{2}+ b_{1}b_{2} }{a^{2}_{2} + b_{2}^{2}} + \frac{ i(a_{2}b_{1} - a_{1}b_{2})}{a^{2}_{2} + b_{2}^{2}}$

Объяснение:

1) $z_{1} = 3 + 7i, z_{2} = -2 - 5i$

$z_{1} + z_{2} = 3 + 7i - 2 - 5i = 1 + 2i$

$z_{1} - z_{2} = 3 + 7i -( - 2 - 5i) = 3 + 7i + 2 + 5i = 5 + 12i$

$z_{1}z_{2} = (3 + 7i)(-2 -5i) = -(3 + 7i)(2 + 5i) = -(6 + 15i + 14i + 35i^{2}) =$

$= -(6 +29i -35) = -(-29 + 29i) = 29 - 29i$

$\displaystyle \frac{z_{1}}{z_{2}} = \frac{3 + 7i}{-2 - 5i} = - \frac{3 + 7i}{2 + 5i} = - \frac{(3 + 7i)(2 - 5i)}{(2 + 5i)(2 - 5i)} = -\frac{6 - 15i + 14i - 35i^{2}}{4 - 25i^{2}} =$

$\displaystyle = -\frac{6 - i + 35}{4 + 25} = -\frac{41 - i}{29} = -\frac{41}{29} + i\frac{1}{29}$

2) $z_{1} = 4 + 3i, z_{2} = 2 - 7i$

$z_{1} + z_{2} = 4 + 3i + 2 - 7i = 6 - 4i$

$z_{1} - z_{2} = 4 + 3i - (2 - 7i) = 4 + 3i - 2 + 7i = 2 + 10i$

$z_{1} z_{2} = (4 + 3i)(2 - 7i) = 8 - 28i + 6i - 21i^{2} = 8 - 22i + 21 = 29 - 22i$

$\displaystyle \frac{z_{1}}{z_{2}} = \frac{4 + 3i}{2 - 7i} = \frac{(4 + 3i)(2 + 7i)}{(2 - 7i)(2 + 7i)} = \frac{8 + 28i + 6i + 21i^{2}}{4 - 49i^{2}} = \frac{8 + 34i - 21}{4 + 49} =$