Задача 75960 Вариант 30 1, Составить квадратное...
Условие
Решение
[m]x^2 - \frac{1-3i}{2} \cdot x - \frac{1+3i}{2} \cdot x + \frac{1-3i}{2} \cdot \frac{1+3i}{2} = 0[/m]
[m]x^2 - \frac{1-3i+ 1+3i}{2} \cdot x + \frac{(1-3i)(1+3i)}{4} = 0[/m]
[m]x^2 - \frac{2}{2} \cdot x + \frac{1+3}{4} = 0[/m]
[m]x^2 - x + 1 = 0[/m]
2) a) (3 - i)(3 + 4i) = 3*3 - 3*i + 3*4i - i*4i = 9 - 3i + 12i + 4 = 13 + 9i
b) [m]\frac{3-i}{4+5i} = \frac{(3-i)(4-5i)}{(4+5i)(4-5i)} = \frac{12-4i-15i+(-5)}{16+25} = \frac{7-19i}{41}[/m]
3) z1 = -2 + 5i; z2 = -5 - 3i
z3 = z1 + z2 = -2 + 5i - 5 - 3i = -7 + 2i
Геометрическая интерпретация - это точки на комплексной плоскости.
Смотрите на рисунке.
4) a) 5(cos 10° + i*sin 10°)*2(cos 80° + i*sin 80°) =
= 10(cos 10°*cos 80° + i*sin 10°*cos 80° + i*cos 10°*sin 80° + i^2*sin 10°*sin 80°) =
= 10(cos 10°*cos 80° - sin 10°*sin 80° + i*(sin 10°*cos 80° + cos 10°*sin 80°)) =
= 10(cos (10° + 80°) + i*sin(10° + 80°)) = 10(cos 90° + i*sin 90°) = 10(0 + i) = 10i
b) [m]\frac{18e^{\pi/2 \cdot i}}{(1-\sqrt{3}i)^2} = \frac{18(cos \frac{\pi}{2} + i \cdot sin \frac{\pi}{2})}{(1-\sqrt{3}i)^2} = \frac{18(0 + i \cdot 1)(1+\sqrt{3}i)^2}{(1-\sqrt{3}i)^2(1+\sqrt{3}i)^2} =[/m]
[m]= \frac{18i(1+2\sqrt{3}i - 3)}{(1+3)^2} = \frac{18i(-2+2\sqrt{3}i)}{16} = \frac{9i(-1+\sqrt{3}i)}{4} = \frac{-9\sqrt{3}-9i}{4}[/m]
5) [m]z=\frac{1-i}{e^{\frac{3\pi}{4} \cdot i}} = \frac{1-i}{cos \frac{3\pi}{4} + i \cdot sin \frac{3\pi}{4}} = \frac{1-i}{-1/\sqrt{2} + i \cdot 1/\sqrt{2}} =[/m]
[m]\frac{(1-i)(- 1/\sqrt{2} - i \cdot 1/\sqrt{2})}{(- 1/\sqrt{2} + i \cdot 1/\sqrt{2})(- 1/\sqrt{2} - i \cdot 1/\sqrt{2})} =\frac{(1-i)(- 1/\sqrt{2} - i \cdot 1/\sqrt{2})}{\frac{1}{2} + \frac{1}{2}} = (1-i)(- \frac{1}{\sqrt{2}} - i \cdot \frac{1}{\sqrt{2}}) =[/m]
[m]=- \frac{1}{\sqrt{2}} - i \cdot \frac{1}{\sqrt{2}} + i \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = - \frac{2}{\sqrt{2}} = -\sqrt{2}[/m]