Apolar polynomial of a polynomial of degree 2 Suppose $P(z)=3z^2-2(a+b+c)z+ab+bc+ca,$ where $a, b, c$ are any complex numbers lying on or within the unit circle $|z|=1.$ I am trying to find an apolar polynomial $Q(z)$ (in terms of a,b,c) to $P(z)$ such that both the zeros of $Q(z)$ lie within a unit distance from  $a.$ Could any one guide me in this direction?

I'm going to charge in
and see how far I can get.
It turns out that
I got stopped,
but I will show how far I got anyway.
The zeros of
$P(z)=3z^2-2(a+b+c)z+ab+bc+ca,
$
(I am assuming that
the second term is $z$, not $z^2$)
are
$\begin{array}\\
r_1, r_2
&=\dfrac{2(a+b+c)\pm \sqrt{4(a+b+c)^2-12(ab+bc+ca)}}{6}\\
&=\dfrac{2(a+b+c)\pm \sqrt{4(a^2+b^2+c^2+2ab+2ac+2bc)-12(ab+bc+ca)}}{6}\\
&=\dfrac{2(a+b+c)\pm \sqrt{4(a^2+b^2+c^2-ab-ac-bc)}}{6}\\
&=\dfrac{(a+b+c)\pm \sqrt{a^2+b^2+c^2-ab-ac-bc}}{3}\\
\end{array}
$
Now
$\begin{array}\\
2(a^2+b^2+c^2-ab-ac-bc)
&=a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\\
&=(a-b)^2+(b-c)^2+(c-a)^2\\
\end{array}
$
so
$r_1, r_2
=\dfrac{(a+b+c)\pm \sqrt{(a-b)^2+(b-c)^2+(c-a)^2)}/\sqrt{2}}{3}
$.
Choosing $a$ as the point
to work with,
$a_1
=a-r_1
=\dfrac{(2a-b-c)- \sqrt{(a-b)^2+(b-c)^2+(c-a)^2)}/\sqrt{2}}{3}
$
and
$a_2
=a-r_2
=\dfrac{(2a-b-c)+ \sqrt{(a-b)^2+(b-c)^2+(c-a)^2)}/\sqrt{2}}{3}
$,
and we want
$|a_1|<1
$
and
$|a_2|<1
$.
Choosing $a_1$,
this means
$|(2a-b-c)+ \sqrt{(a-b)^2+(b-c)^2+(c-a)^2)}/\sqrt{2}|
< 3
$.
The fact that
$a, b,$ and $c$
are complex
leaves me unsure of what to do
at this point.
So,
I'll stop and hope that
someone else can make use of
what I have done.
