# Equal absolute values of roots of Quadratic equation

Show that for quadratic equation $$z^2+az+b^2=0 \quad a,b\in\mathbb{C}$$ which roots $$z_1,z_2$$ has the property that $$|z_1| = |z_2|$$ we can observe that $$\frac{a}{b}\in\mathbb{R}$$

Not full solution:

What I can observe is that if $$|z_1| = |z_2|$$ then $$|z_1|^2 = |z_2|^2$$ so $$z_1\overline{z_1}=z_2\overline{z_2}$$ so lets say $$z_s=z_1\overline{z_1}=z_2\overline{z_2}$$ and using Vieta's formulas we can see that $$z_1+z_2=-a\quad\quad\quad\quad\quad\quad z_1z_2=b^2$$ so $$\overline{z_1}+\overline{z_2}=-\overline{a}\quad\quad\quad\quad\quad\quad\overline{z_1}\overline{z_2}=\overline{b}^2$$ and by combining these values we get $$a\overline{b}^2=-(z_1+z_2)\overline{z_1}\overline{z_2}=-z_s(\overline{z_1}+\overline{z_2})=z_s\overline{a}$$ $$\overline{a}b^2=-(\overline{z_1}+\overline{z_2})z_1z_2=-z_s(z_1+z_2)=z_s a$$ so we see that $$\frac{a}{\overline{a}}\overline{b}^2 = z_s = \frac{\overline{a}}{a}b^2$$ and so $$\frac{a^2}{b^2} = \frac{\bar{a}^2}{\bar{b}^2}$$ so or $$\frac{a}{b}-\frac{\bar{a}}{\bar{b}}=0$$ and then $$\operatorname{Im}\frac{a}{b}=0$$ or $$\frac{a}{b}+\frac{\bar{a}}{\bar{b}}=0$$ and then $$\operatorname{Re}\frac{a}{b}=0$$

And at the end I don't know why the second option isn't possible so I cannot prove that only the first one is possible.

I recommend casting the complex numbers $$z_1$$, $$z_2$$ in polar form, to wit:

With $$\vert z_1 \vert = \vert z_2 \vert$$, we may write

$$z_1 = re^{i\theta}, \; z_2 = re^{i\phi}; \tag 1$$

now, $$z_1$$, $$z_2$$ being roots of

$$z^2 + az + b^2 = 0 \tag 2$$

implies

$$(z - z_1)(z - z_2) = z^2 - (z_1 + z_2)z + z_1 z_2 = z^2 + az + b^2 = 0; \tag 3$$

thus

$$a = -(z_1 + z_2), \; b = z_1 z_2; \tag 4$$

using (1),

$$a = -r(e^{i\theta} + e^{i\phi}), \; b^2 = r^2 e^{i(\theta + \phi)}; \tag 5$$

$$b = \pm re^{i(\theta + \phi)/2}; \tag 6$$

$$\dfrac{a}{b} = \pm \dfrac{e^{i\theta} + e^{i\phi}}{e^{i(\theta + \phi)/2}} = \pm e^{-i(\theta + \phi)/2}(e^{i\theta} + e^{i\phi}) = \pm(e^{i(\theta - \phi)/2} + e^{i(\phi - \theta)/2}); \tag 7$$

we close by simply observing that

$$e^{i(\phi - \theta)/2} = e^{-i(\theta - \phi)/2} = \overline{e^{i(\theta - \phi)/2}}, \tag 8$$

and thus (7) becomes

$$\dfrac{a}{b} = \pm(e^{i(\theta - \phi)/2} + \overline{e^{i(\theta - \phi)/2}}) = \pm 2 \cos \left ( \dfrac{\theta - \phi}{2} \right ) \in \Bbb R. \tag 9$$

Hint:

Note that saying $$\frac{a^2}{b^2}=\frac{\bar a^2}{\bar b^2}$$ is the same as

$$\left(\frac{a^2}{b^2}\right)=\overline{\left(\frac{a^2}{b^2}\right)}\tag{1}$$ That is, a complex number is equal to it's conjugate.

Letting $$x+yi=\frac{a}{b}$$, you now know that $$(x+iy)^2=x^2-y^2+2xyi$$ is real. What can you conclude about $$x,y$$?

• So I know that $\frac{a^2}{b^2}$ is real number, but then $\frac{a}{b}$ in my opinion doesn't have to be real number – user608151 Oct 24 '18 at 21:45
• @MaciejProcyk Well, in your post you said $a,b\in\Bbb C$. Try rewriting $\left(\frac{a^2}{b^2}\right)=x+iy$, $x,y\in\Bbb R$, and look at $(1)$. – cansomeonehelpmeout Oct 24 '18 at 21:46
• So $x+0i=\frac{a^2}{b^2}$? – user608151 Oct 24 '18 at 21:58
• @MaciejProcyk Yes! – cansomeonehelpmeout Oct 24 '18 at 21:59
• So $\frac{a^2}{b^2}$ is real number but I wanted to show that $\frac{a}{b}$ is real number – user608151 Oct 24 '18 at 22:00