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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.

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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

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

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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 $x,y$?

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  • $\begingroup$ 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 $\endgroup$ – user608151 Oct 24 '18 at 21:45
  • $\begingroup$ @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 $(1)$. $\endgroup$ – cansomeonehelpmeout Oct 24 '18 at 21:46
  • $\begingroup$ So $x+0i=\frac{a^2}{b^2}$? $\endgroup$ – user608151 Oct 24 '18 at 21:58
  • $\begingroup$ @MaciejProcyk Yes! $\endgroup$ – cansomeonehelpmeout Oct 24 '18 at 21:59
  • $\begingroup$ So $\frac{a^2}{b^2}$ is real number but I wanted to show that $\frac{a}{b}$ is real number $\endgroup$ – user608151 Oct 24 '18 at 22:00

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