x1 = -re(acos(-3)) + 2*pi - i*im(acos(-3))
x2 = 2*pi - i*im(acos(4))
x3 = re(acos(-3)) + i*im(acos(-3))
x4 = re(acos(4)) + i*im(acos(4))
Объяснение:
x1 = -re(acos(-3)) + 2*pi - i*im(acos(-3))
x2 = 2*pi - i*im(acos(4))
x3 = re(acos(-3)) + i*im(acos(-3))
x4 = re(acos(4)) + i*im(acos(4))
x1 = 3.14159265358979 + 1.76274717403909*i
x2 = 6.28318530717959 - 2.06343706889556*i
x3 = 3.14159265358979 - 1.76274717403909*i
x4 = 2.06343706889556*i
сумма
-re(acos(-3)) + 2*pi - i*im(acos(-3)) + 2*pi - i*im(acos(4)) + i*im(acos(-3)) + re(acos(-3)) + i*im(acos(4)) + re(acos(4))
=
4*pi + re(acos(4))
произведение
(((-re(acos(-3)) + 2*pi - i*im(acos(-3)))*(2*pi - i*im(acos(4*(i*im(acos(-3)) + re(acos(-3*(i*im(acos(4)) + re(acos(4)))
=
-(2*pi - i*im(acos(4)))*(i*im(acos(-3)) + re(acos(-3)))*(i*im(acos(4)) + re(acos(4)))*(-2*pi + i*im(acos(-3)) + re(acos(-3)))
x-x1 y-y1
= x1=-1 x2=3 y1=8 y2=-4
x2-x1 y2-y1
x-(-1) y-8 x+1 y-8 x+1 y-8
= ⇔ = или =
3-(-1) -4-8 4 -12 1 -3
-3(x+1)=y-8 или y=-3x+5
y=kx+b
A(-1;8) ∈ y=kx+b ⇔ 8=k(-1)+b -k+b=8
и B(3;-4)∈ y=kx+b ⇔-4=k(3)+b ⇔ 3k+b=-4 ⇔4k=-12 k=-3
b=8+k=5
y=-3x+5
проверка
A(-1;8) и B(3;-4)∈ y=kx+b y=-3x+5
A(-1;8) 8=-3(-1)+5 верно
B(3;-4) -4=-3(3)+5 верно