Show a complex equation has one or two roots Let $a$ $\neq$ $0$, $b,$ and $c$ be complex constants.  Show that the quadratic equation $az^2+bz+c=0$ has one or two roots.
My thoughts:
Let $a=a_1+ia_2,$ $b=b_1+ib_2,$ and $c=c_1+ic_2$.
I also think I need to use $\frac{-b\pm\sqrt{b^2+4ac}}{2a}$, where $a$ $\neq$ $0$.
By plugging the above in, I get:
$$(a_1+ia_2)z^2+(b_1+ib_2)z+c_1+ic_2=0$$
$$\Downarrow$$
$$a_1z^2+ia_2z^2+b_1z+ib_2z+c_1+ic_2=0$$
$$\Downarrow$$
$$a_1z^2+b_1z+c_1+i(a_2z^2+b_2z+c_2)=0$$
And now I'm stuck and not sure what to do next.  Any help would be appreciated!
 A: You're going about it the wrong way. Try this:


*

*Divide through by $a$, to get an equation of the form $z^2+Bz+C = 0$, which has the same roots as the original equation. (This step is not strictly necessary, but it simplifies the algebra.)

*Take the two roots $\alpha, \beta$ that you know about (if the equation has only one root $\alpha$, set $\beta = \alpha$). Show (by direct algebraic calculation) that $(z-\alpha)(z-\beta) = z^2+Bz+C$.

*Deduce that $z^2+Bz+C$ can't be zero unless $z=\alpha$ or $z=\beta$.

A: Continuing your equation, we'd get that the Real part = 0 and Imaginary part = 0. Should be easy, knowing that your $a_1, b_1, c_1, a_2, b_2,  c_2$ are real numbers
A: If it has at least two zeros, then $az^2+bz+c-a(z-z_1)(z-z_2)$ also has two zeros, but it is at most a linear function $dz+e$.  So $z_1=-e/d=z_2$ unless $d=e=0$.  So $az^2+bz+c=a(z-z_1)(z-z_2)$.  If $az_3^2+bz_3+c=0$ then $a(z_3-z_1)(z_3-z_2)=0$ so $z_3=z_1$ or $z_2$.
As for how to show it has at least one zero, I'd ask Gauss or someone.
