Images: An example of finding explicit equations for curves using complex functions that make up an implicitly defined cubic curve

It should be noted that -The curve of the implicit initial cubic equation is generally composed of 3 known exact solutions - complex functions: f1(z), f2(z) and f3(z). Each of them consists of a real and an imaginary part: f(z)=Re(f(z))+ί*Im(f(z)). -The parts of these curves, which form the initial cubic curve of the implicit function, correspond to the regions in the complex plane: ΔZ=[z1,z2], where their imaginary parts are absent: Im(f(z))=0 for z∈ΔZ. Images made with applet.
Example #2 eq: 8.3 y³ - 2.2 y² - 39.6 y - 5 x³ + x² + 45x - 14.3 =0

f1(z)

 f1(z)

f2(z)

f2(z)

f3(z)

f3(z)

Example #2 The resulting representation of a function given by an implicit formula: eq: 8.3 y³ - 2.2 y² - 39.6 y - 5 x³ + x² + 45x - 14.3 =0

  Example #2   The resulting representation of a function given by an implicit formula: eq: 8.3 y³ - 2.2 y² - 39.6 y - 5 x³ + x² + 45x - 14.3 =0
The resulting representation of the function given by the implicit formula consists of three branches.
Example #1 eq: -4 y³ + 4.2 y² + 36 y - 5 x³ + x² + 45x = 0

The resulting representation of the function given by the implicit formula consists of three branches.
Example #3 eq;: -5 y³ - 2.2 y² - 10.5 y - 5 x³ + x² + 45x - 18.7 =0

The solution of the third-degree Polynomial equation consists of three complex functions: f1(z), f2(z), f3(z). Since the imaginary part of the complex functions f2(z) and f3(z) along the entire real axis is different from zero, except for f1(z), then the plane curve consists only of it Cu1=Curve(a, If(imaginary(f1(a + 0ί)) ≠ 0, 6000, real(f1(a + 0ί))), a, x(Corner(1, 1)), x(Corner(1, 2))), which has its own domain of definition. The resulting representation of the function given by the implicit formula consists of one branch.