Упр.586 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

Найти произведение комплексных чисел:

1. $$\left(2+3i\right)\left(4+5i\right)$$;
2. $$\left(1-2i\right)\left(5-i\right)$$;
3. $$\left(-3-2i\right)\left(2+3i\right)$$;
4. $$\left(3-i\right)\left(3+i\right)$$;
5. $$\left(-\frac{1}{2}+1{,}5i\right)\left(2-4i\right)$$;
6. $$\left(\sqrt{2}+3i\right)\left(\sqrt{2}+3i\right)$$.

1. $$\begin{aligned} (2+3i)(4+5i)&=2\cdot4+2\cdot5i+3i\cdot4+3i\cdot5i\\ &=8+10i+12i+15i^2\\ &=8+22i-15\\ &=-7+22i \end{aligned}$$
2. $$\begin{aligned} (1-2i)(5-i)&=1\cdot5+1\cdot(-i)+(-2i)\cdot5+(-2i)\cdot(-i)\\ &=5-i-10i+2i^2\\ &=5-11i-2\\ &=3-11i \end{aligned}$$
3. $$\begin{aligned} (-3-2i)(2+3i)&=(-3)\cdot2+(-3)\cdot3i+(-2i)\cdot2+(-2i)\cdot3i\\ &=-6-9i-4i-6i^2\\ &=-6-13i+6\\ &=-13i \end{aligned}$$
4. $$\begin{aligned} (3-i)(3+i)&=3\cdot3+3\cdot i+(-i)\cdot3+(-i)\cdot i\\ &=9+3i-3i-i^2\\ &=9+1\\ &=10 \end{aligned}$$
5. $$\begin{aligned} \left(-\frac12+1{,}5i\right)(2-4i)&=-\frac12\cdot2-\frac12\cdot(-4i)+1{,}5i\cdot2+1{,}5i\cdot(-4i)\\ &=-1+2i+3i-6i^2\\ &=-1+5i+6\\ &=5+5i \end{aligned}$$
6. $$\begin{aligned} (\sqrt2+3i)(\sqrt2+3i)&=(\sqrt2)^2+\sqrt2\cdot3i+3i\cdot\sqrt2+(3i)^2\\ &=2+3\sqrt2\,i+3\sqrt2\,i+9i^2\\ &=2+6\sqrt2\,i-9\\ &=-7+6\sqrt2\,i \end{aligned}$$

Ответ

1) $$-7+22i$$; 2) $$3-11i$$; 3) $$-13i$$; 4) $$10$$; 5) $$5+5i$$; 6) $$-7+6\sqrt2\,i$$.