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 $w_2=\frac{1}{\overline{w_1}}$
$\implies w_1\overline {w_2}=1$

Worst comes to worst, you can bash this. Let $a+\overline{c} = t$ and $b+\overline{b} = 2s$ with $s$ real. Then your quadratic is:
$$tw^2+2sw+\overline{t} = 0\implies w_{1,2} = \dfrac{-s\pm\sqrt{s^2-|t|^2}}{t}.$$
Now, observe that the entry inside the square root in the discriminant is real. This means that when you compute the norm of the numerator of $w_1$, (with the choice of plus sign in the middle):
$$\left|-s+\sqrt{s^2-|t|^2}\right|^2 = (-s+\sqrt{s^2-|t|^2})(-s-\sqrt{s^2-|t|^2}) = |t|^2$$
and so $|w_1| = 1.$ Can do the same for $|w_2| = 1.$
Key observation here is that $s$ is real and so is $s^2-|t|^2,$ which enables us to compute the norm squared as itself times its conjugate. Lastly, $|z_1|, |z_2| < 1$ ensures $t\neq 0$ from Vieta.
