Пусть комплексное число имеет вид: z = x + iy
Модуль комплексного числа: ![|z|=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}](https://tex.z-dn.net/?f=%7Cz%7C%3D%5Csqrt%7B%28-3%29%5E2%2B%28-2%29%5E2%7D%3D%5Csqrt%7B13%7D)
Так как x , y < 0, то угол α ∈ III четверти, тогда
![\alpha=\rm \pi+arctg\bigg|\dfrac{y}{x}\bigg|=\pi+arctg\dfrac{2}{3}](https://tex.z-dn.net/?f=%5Calpha%3D%5Crm+%5Cpi%2Barctg%5Cbigg%7C%5Cdfrac%7By%7D%7Bx%7D%5Cbigg%7C%3D%5Cpi%2Barctg%5Cdfrac%7B2%7D%7B3%7D)
![\rm z=-3-2i=\sqrt{13}\left[\cos\bigg(arctg\dfrac{2}{3}+\pi\bigg)+i\sin\bigg(arctg\dfrac{2}{3}+\pi\bigg)\right]](https://tex.z-dn.net/?f=%5Crm+z%3D-3-2i%3D%5Csqrt%7B13%7D%5Cleft%5B%5Ccos%5Cbigg%28arctg%5Cdfrac%7B2%7D%7B3%7D%2B%5Cpi%5Cbigg%29%2Bi%5Csin%5Cbigg%28arctg%5Cdfrac%7B2%7D%7B3%7D%2B%5Cpi%5Cbigg%29%5Cright%5D)
Решение
- 2 ≤ -1,2 х - 3 ≤ 3
- 2 + 3 ≤ -1,2 х <span>≤ 3 + 3
</span>1 ≤ -1,2 х <span>≤ 6
</span>6/(- 1,2) ≤ x ≤ 1/(- 1,2)
- 5 <span>≤ x ≤ - 5/6
</span>x ∈ [ - 5 ; - 5/6]
.................................
128+128+129+130+132+135+135+137+142=1196
1196÷9=132.8средний рост
3)4