Let $z_1$ and $z_2$ be the roots of $az^2+bz+c=0;\,a,b,c\in \mathbb{C}, a\ne 0$ and $w_1,w_2$ be roots of $(a+\bar{c})z^2+(b+\bar{b})z+(\bar{a}+c)=0$ Let $z_1$ and $z_2$ be the roots of $az^2+bz+c=0;\,\,a,b,c\in \mathbb{C}, a\ne 0$ and $w_1,w_2$ be roots of $(a+\overline{c})z^2+(b+\overline{b})z+(\overline{a}+c)=0$. If $|z_1|<1,|z_2|<1$, then prove that $|w_1|=|w_2|=1$
My Progress:
We can see that whenever $z$ is a root of $(a+\overline{c})z^2+(b+\overline{b})z+(\overline{a}+c)=0,\,\, \frac{1}{\overline{z}}$ is also a root.
Case 1: $w_1=\frac{1}{\overline{w_1}}$ and $w_2=\frac{1}{\overline{w_2}}$
$\implies |w_1|=|w_2|=1$
Case 2: $w_1=\frac{1}{\overline{w_2}}$ and $w_2=\frac{1}{\overline{w_1}}$
$\implies w_1\overline {w_2}=1$

Let $a+\overline{c} = t$ and $b+\overline{b} = 2s$ with $s$ real.
Claim: $t\neq 0\text{ and }s^2-|t|^2 <0$
proof:
(i)  $t=0\implies a=-\bar{c}\implies |z_1z_2|=1$ contradiction!!
(ii)  let $\frac{\overline{a}}{a}=\alpha.$
\begin{align*}
4(s^2-|t|^2)&= |a(z_1+z_2)+\overline{a(z_1+z_2)}|^2-4|a+\overline{az_1z_2}|^2\\
&=-|a|^2[4+4|z_1|^2|z_2|^2-2|z_1+z_2|^2-2\Re{(\overline\alpha(z_1-z_2)^2)}]\\
&\le-|a|^2[4+4|z_1|^2|z_2|^2-2|z_1+z_2|^2-2|\overline\alpha||z_1-z_2|^2]\\
&=-4|a|^2(1-|z_1|^2)(1-|z_2|^2)\\ 
&\le 0
\end{align*}
Now
\begin{align*}
tw^2+2sw+\overline{t} &= 0\\
\implies w_{1,2} &= \dfrac{-s\pm\sqrt{s^2-|t|^2}}{t}\\
\implies w_{1,2} &= \dfrac{-s\pm\iota\sqrt{|t|^2-s^2}}{t}\\
\implies |w_{1,2}| &= \dfrac{\sqrt{|-s|^2+|t|^2-s^2}}{|t|}\\
&=1
\end{align*}
