Show a complex equation has one or two roots

Question

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!

Answer 1

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

Answer 2

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.

Answer 3

You're going about it the wrong way. Try this:

1. 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.)
2. 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 $(z-\alpha)(z-\beta) = z^2+Bz+C$.
3. Deduce that $z^2+Bz+C$ can't be zero unless $z=\alpha$ or $z=\beta$.