Proving $\overline{a}b|c|=|a|\overline{b}c$ for equation $az^2+bz+c=0$

Question

If $a,b,c \in \mathbb{C},\ a\neq0$ such that the equation $az^2+bz+c=0$ has roots with equal modules, prove that

$$\overline{a}b|c|=|a|\overline{b}c$$

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I tried to let $z_1,\ z_2$ the roots with $|z_1|=|z_2|=k$ and from Vieta $z_1+z_2=-\frac{b}{a}$ and $z_1z_2=\frac{c}{a}$. From the latter $k^2|a|=|c|$. So I was thinking, it is enough to prove: $\overline{a}bk^2=\overline{b}c$, but I don't how to use the first equation from Vieta and I am stuck.

Answer 1

Let $$ z_1=re^{i\theta_1}, z_2=re^{i\theta_2} $$ where $r>0$ and $\theta_1,\theta_2$ are real. By Vieta's Theorem, $$ re^{i\theta_1}+re^{i\theta_2}=-\frac{b}{a}, r^2e^{i(\theta_1+\theta_2)}=\frac{c}{a} $$ which gives $$ b=-ar(e^{i\theta_1}+e^{i\theta_2}), c=ar^2e^{i(\theta_1+\theta_2)} $$ So $$ \overline{a}b|c|=\overline{a}\bigg[-ar(e^{i\theta_1}+e^{i\theta_2})\bigg]|a|r^2=-|a|^3r^3(e^{i\theta_1}+e^{i\theta_2})$$ and $$ |a|\overline{b}c=|a|\bigg[-\bar{a}r(e^{-i\theta_1}+e^{-i\theta_2})\bigg]ar^2e^{i(\theta_1+\theta_2)}=-|a|^3r^3(e^{i\theta_1}+e^{i\theta_2}) $$ and hence $$ \overline{a}b|c|=|a|\overline{b}c. $$

Answer 2

Given $az^2+bz+c=0$, we have

$$z_1+z_2 = -\frac ba, \>\>\>\>\>z_1z_2 = \frac ca$$

From the equal module $|z_1|=|z_2|=k$, we have

$$k^2=|z_1z_2|=\frac{|c|}{|a|}$$

Then, multiply , $\bar z_1+\bar z_2 = -\frac {\bar b}{\bar a}$ and $z_1z_2 = \frac ca$ to get

$$- \frac {\bar b c}{| a|^2}=(\bar z_1+\bar z_2)z_1z_2 = (z_1+z_2 )k^2 =-\frac ba \frac{|c|}{|a|}$$

Rearrange to obtain,

$$\overline{a}b|c|=|a|\overline{b}c$$