Question about quadratic equation of complex coefficients.

Question

Let $az^2+bz+c=0$ be a quadratic equation with complex coefficients $a,b,c$ and roots $z_1, z_2.$

If it is given that $|z_1|\not=|z_2|,$ how can I obtain the condition for this containing $a,b,c?$

Is there any reference discuss about roots of quadratic equations with complex coefficients?

The classical formulas hold for complex coefficients as well. — Yves Daoust, Oct 22 '14 at 7:48
@YvesDaoust: But it is hard to obtain condition on $|z_1|,|z_2|$ using that formula. — Bumblebee, Oct 22 '14 at 8:00
We know that $a\not=0.$ Therefore $||z_1|-|z_2||\le|z_1+z_2|=|-\dfrac{b}{a}|.$ Implies "the necessary condition" would be $b\not=0.$ But it is not sufficient. — Bumblebee, Oct 22 '14 at 8:58

Yves Daoust · Accepted Answer · 2014-10-22 09:04:24Z

Let us express that the magnitudes are equal. Using the classical formula:

$$\left|\frac{-b+\sqrt{b^2-4ac}}{2a}\right|=\left|\frac{-b-\sqrt{b^2-4ac}}{2a}\right|.$$

If $b=0$, the equality holds.

Otherwise, we can multiply by $|-2a/b|$, and

$$\left|1-\sqrt{1-4\frac{ac}{b^2}}\right|=\left|1+\sqrt{1-4\frac{ac}{b^2}}\right|.$$

Now, $|1-p|=|1+p|\implies \Re(p)=0$. Indeed, $(1-x)^2+y^2=(1+x)^2+y^2\implies x=0$.

So $$\sqrt{1-4\frac{ac}{b^2}}=iy,$$ $$ac=\frac{1-y^2}4b^2.$$

In conclusion, the magnitudes differ if $b$ is nonzero and $ac$ is not the product of $b^2$ by a real number less than or equal to $1/4$.

$\begingroup$ Wonderfull. I got it. $\endgroup$ – Bumblebee Oct 23 '14 at 7:52 — Bumblebee, Oct 23 '14 at 7:52

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