x = (-b+sqrt(b^2-4ac))/(2a) OR x = (-b-sqrt(b^2-4ac))/(2a)
In particular what happens in the following cases of the quadratic formula?
- a=1, b=1
- b=1, c=1
- a=1, c=1
When you merge the images for a=1, c=1 for the positive and negative version of quadratic formula, you get a hyperbola in the real plane and an ellipse in the complex plane.
I understand that we are just plotting the possible conic sections for ax^2+bx+c=0, i.e. ellipses, hyperbolas, parabolas, lines. Is that also the explanation for these stray straight lines in these diagrams or is there another explanation for that?



It's more interesting to plot both solutions together
There are no "stray lines" here; plotting in three dimensions, the graph is the union of
The real plot is looking from above, and thus sees the ellipse edge-on. The imaginary plot is looking from the side, and thus sees the hyperbola edge on.
Anyways, even this appearance is an artifact, due to the restriction to real b. We could shift slightly to an imaginary part of 0.1, and the graphs become
Or even further to an imaginary part of 1:
Ultimately, the figure is a conic section in complex two-space carved out by the equation $z^2 + bz + 1 = 0$.
In real dimensions, that is a two-dimensional surface embedded in four-dimensional space. And it a smooth surface; it has no singular points when viewed in its geometrically correct setting.
Instead, this weird transition from a hyperbola to an ellipse is simply an artifact of additionally adding in the equation $\mathrm{Im}(b) = 0$