Which two roots of this quartic polynomial lie within the unit disk?

I have the following quartic polynomial,

$$f(z) = z^4-rz^3+rsz^2-rz+1$$, where $r\in\mathbb{R}$ and $s\in\mathbb{C}$.

Since this polynomial is palindromic, the roots can be easily computed. They are given by

$$ u_{\pm} = \frac{c_+ \pm \sqrt{c_+^2-4}}{2}, v_{\pm} = \frac{c_{-} \pm \sqrt{c_{-}^2-4}}{2} $$ where, $$ c_{\pm} = \frac{-r\pm\sqrt{r^2-4(rs-2)}}{2}. $$ Another consequence of this polynomial being palindromic is that $f(w)=0$ implies $f(1/w)=0$. This means that either $|u_+|\leq1$ or $|u_-|\leq1$ and one has the same condition for $v_{\pm}$.

My question is: How do I determine which two roots will lie inside the unit disk for a given $s$ and $r$?

My strategy to answer the question has been to compute the modulus squared of each root but I haven't made much progress(currently have lengthy expressions that have to be simplified and then I have to see when they are less than 1).

The two pairs of roots are (after a minor correction) of the form $\displaystyle\,w_\pm = \frac{c \pm \sqrt{c^2-4}}{2}\,$. As noted, $\,w_+\cdot\,w_-=1\,$, so one of the roots is always inside the unit disk (or on its boundary, the unit circle), and that root must be the one with the smaller modulus between the two.

If $\,c \in (-2,2)\,$ then $\,|w_+|=|w_-|=1\,$, otherwise $\,z=c^2-4 \not\in \mathbb R^-\,$ and $\,\overline{\sqrt{z}}=\sqrt{\,\overline{z}\,}\,$ holds, so:

$$ 4\,|w_\pm|^2 = \left(c \pm \sqrt{c^2-4}\right)\left(\bar c \pm \sqrt{\bar c^2-4}\right) = |c|^2 + |c^2-4| \pm 2 \text{Re}\left(\bar c \sqrt{c^2-4}\right) $$

It follows that the smaller modulus is the one for which the last term is negative, which is $\,w_-\,$ if $\,\text{Re}\left(\bar c \sqrt{c^2-4}\right) \ge 0\,$, or $w_+$ otherwise.