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 $(a+\bar{c})z^2+(b+\bar{b})z+(\bar{a}+c)=0$

Question

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

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

Answer 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*}

Answer 2

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.