Every quadratic polynomial is conjugate to one of the polynomial $z^2+t$
where conjugation means conjugation by a linear fractional transformation $\alpha(z)=\frac{az+b}{cz+d}$ and $f^{\alpha}={\alpha}^{-1}f{\alpha}$.
Now if $p$ is said to be a fixed point by $f$ then ${\alpha}^{-1}(p)$ is said to be a fixed point by $f^{\alpha}$.
Again any quadratic polynomial can have two distinct or same root in $\Bbb C$. I think I have to connect this but I can't! Even in mathstack I got this but not what I want.
I think I got one idea if we just consider the affine linear transformation ${\alpha}(z)=az+b$ then
If $f$ has two distinct roots say $x,y\in \Bbb C$ then we choose $t\neq 0$ then we can choose ${\alpha}(z)$ s.t ${\alpha}(x)=t$ and ${\alpha}(y)=-t$ then $f\circ {\alpha}^{-1}(t)=0$ and $f\circ {\alpha}^{-1}(-t)=0$. Moreover, $f\circ {\alpha}^{-1}(z)\to \infty$ as $z \to \infty$. Can we say something using these?
This is basically just completing the square. Every quadratic has the form $a(z+b)^2+c$ for some $a,b,c\in\mathbb{C}$ with $a$ nonzero. Conjugating by $z\mapsto z-b$ you get $az^2+d$ for some $d$ and then conjugating by $z\mapsto z/a$ you get $z^2+t$ for some $t$.