<s>[INST] Visualization of complex roots for quadratics I read that if a parabola has no real roots, then its complex roots can be visualized by graphing the same parabola ($ax^2 + bx + c$) with $-a$ and then finding the roots of that, then using those roots as the diameter of a circle. Then, the "top" and "bottom" of the circle will be the complex roots of the parabola (in the complex plane). 
I don't know if I explained that in [enough] detail, but I'm trying to prove this. I've been trying to generalize an equation for all the parabolas with no real roots, and I seem to be getting somewhere, but it seems like I'm over-complicating this. 
How would you approach this? Any suggestions would be appreciated. 
A proof of this construction would also be an acceptable answer. Thank you.
 [/INST] Either I'm misunderstanding what you wrote, I did a mistake below, or the construction does not work.
Assume that $a\gt 0$ (we may as well, since the roots of $ax^2+bx+c$ are the same as the roots of $-ax^2-bx^2-c$).
The roots of $-ax^2+bx+c$ are the same as the roots of $ax^2-bx-c$, and are given by
$$\frac{b+\sqrt{b^2+4ac}}{2a}\quad\text{and}\quad\frac{b-\sqrt{b^2+4ac}}{2a}.$$
If you are assuming that $ax^2+bx+c$ has no real roots (so that $b^2-4ac\lt 0$), then this has real roots, since $b^2-4ac\lt 0$ implies that $ac\gt 0$, so $b^2+4ac\gt 0$.
The circle whose diameter goes from the point $\left(\frac{b-\sqrt{b^2+4ac}}{2a},0\right)$ to $\left(\frac{b+\sqrt{b^2+4ac}}{2a}\right)$ has center at $\left(\frac{b}{2a},0\right)$, and radius $\frac{\sqrt{b^2+4ac}}{2a}$. So the coordinates of the "top" and "bottom" of that circle are
$$\left(\frac{b}{2a},\frac{\sqrt{b^2+4ac}}{2a}\right)\quad\text{and}\quad\left(\frac{b}{2a},-\frac{\sqrt{b^2+4ac}}{2a}\right).$$
That is, they correspond to the complex numbers
$$\frac{b}{2a} + \frac{\sqrt{b^2+4ac}}{2a}i\qquad\text{and}\qquad \frac{b}{2a}-\frac{\sqrt{b^2+4ac}}{2a}i.$$
These are the roots of $ax^2+bx+c=0$ if and only if they are the roots of $x^2+\frac{b}{a}x + \frac{c}{a}=0$, if and only if they add up to $-\frac{b}{a}$ and multiply to $\frac{c}{a}$. But they add up to $\frac{b}{a}$, not $-\frac{b}{a}$; and their product is
$$\left(\frac{b}{2a} + \frac{\sqrt{b^2+4ac}}{2a}i\right)\left(\frac{b}{2a}-\frac{\sqrt{b^2+4ac}}{2a}i\right) = \frac{b^2}{4a^2} + \frac{b^2+4ac}{4a^2} = \frac{b^2+2ac}{2a^2}\neq \frac{c}{a}.$$
For an explicitly example, take $x^2+x+1$, which has no real roots; the complex roots are $\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\frac{1}{2}-\frac{\sqrt{3}}{2}i$. The construction you describe begins by considering instead $-x^2+x+1$, whose roots are $\frac{1}{2}+\frac{\sqrt{5}}{2}$ and $\frac{1}{2}-\frac{\sqrt{5}}{2}$. The circle whose diameter goes from $(\frac{1}{2}-\frac{\sqrt{5}}{2},0)$ to $(\frac{1}{2}+\frac{\sqrt{5}}{2},0)$ has center at $(\frac{1}{2},0)$ and radius $\frac{\sqrt{5}}{2}$, so the "top" and "bottom" of the circle will be $(\frac{1}{2},\frac{\sqrt{5}}{2})$ and $(\frac{1}{2},-\frac{\sqrt{5}}{2})$, which are not the complex roots of $x^2+x+1$.

The construction works if $b=0$: in that case, instead of graphing $ax^2+c$, with $\frac{c}{a}\gt 0$, we graph $-ax^2+c$; the zeros are $\pm\sqrt{\frac{c}{a}}$, so the circle in question will be centered at the origin and have radius $\sqrt{\frac{c}{a}}$, so the complex numbers corresponding to the "top" and "bottom" are exactly $i\sqrt{\frac{c}{a}}$ and $-i\sqrt{\frac{c}{a}}$, which are the roots of $ax^2+c$ when $\frac{c}{a}\gt 0$. 
So this suggests that for the general case you should first complete the square and then change the sign of the entire squared factor. For my example, beginning with $x^2+x+1$, first complete the square:
$$x^2 + x + 1 = \left(x^2 + x + \frac{1}{4}\right) + \frac{3}{4} = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}.$$
Then graph $-(x+\frac{1}{2})^2 + \frac{3}{4} = -x^2 -x + \frac{1}{2}$ and proceed as you describe. This will work: it amounts to first doing a horizontal shift so that we are dealing with a quadratic of the form $Az^2 + C$ (where $z=x+k$ for some $k$), and then proceeding as above. But you'll note that the resulting equation is not obtained simply by changing the sign of $a$: the values of $b$ and $c$ are also changed.
You can check that this modified idea works by doing essentially what I did above, but showing it does work.
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