[size=85][u][b]Statement of the problem:[/b][/u][br]The curve is given in [b]implicit[/b] form as g(x, y) = 0, for example a circle as y² + x² = 1.[br][u][b] Find:[/b][/u][br]the [b]explicit [/b]form of the equations y=f(x) for each of the [b]k[/b] branches of the curve, i.e. {f[b]ᵢ[/b](x)}, where i=1..k. For a circle, for example, f₁(x) = sqrt(1-x²) for y > 0, and f₂(x) = − sqrt(1-x²) for y < 0. If f(x, y) is a quadratic function with respect to variable [b]y[/b], GeoGebra can easily find the [b]roots [/b]in symbolic form, y₁(x) and y₂(x).[br] For polynomials of the 3rd and 4th degree, knowing the existing rigorous [url=https://www.geogebra.org/m/v4fvf8nx][u][b]solutions[/b][/u][/url] of their equations in symbolic form, one can find the equations of the branches {f[b]ᵢ[/b](x)} of the corresponding plane curves using complex functions.[br] Earlier, I considered biquadratic equations for the [url=https://www.geogebra.org/m/kwdbdznt][u][b]Trifolium curve[/b][/u][/url] and the [url=https://www.geogebra.org/m/cgfbcxau][u][b]Cartesian oval[/b][/u][/url]. These examples illustrate the application of complex functions. In these cases, it is possible to find the equations of the branches of curves using both [i][b]real functions[/b][/i] and [i][b]complex functions[/b][/i], from the real part of the functions of which only those sections are left where the corresponding complex part is missing, i.e. Im(x)=0.[br][/size][size=85] This applet considers an example from mathstackexchange: [url=https://math.stackexchange.com/questions/3980344/branches-of-implicitly-defined-cubic-curves]Branches of implicitly defined cubic curves[/url] for an implicitly defined cubic curve [b]a y³ + b y² + c y + d(x) = 0 where d(x)=-5 x³ + x² + 45x - c0[/b], for which I found explicit equations of the curves of its branches using complex functions.[br] Images of three examples are in the [url=https://www.geogebra.org/m/rek5mz4n]applet.[/url][/size]