For any integer $q>5$ such that $5\nmid q$, set $f_q(x)=x^q-20x^5-1$. All the roots of this polynomial are simple and so it has $q$ distinct roots $\alpha_1,\ldots, \alpha_q$.
By using Mathematica software I observed the following pattern. In fact, one can split $$ \{\alpha_1,\ldots, \alpha_q\}=\{\beta_1,\ldots, \beta_{q-5}\}\cup \{\gamma_1,\ldots, \gamma_5\} $$ such that, $|\beta_i|$ tends to $1$ as $q\to \infty$ (for all $i\in [1,q-5]$) and $|\gamma_j|$ tends to the $j$-th fifth root of $1/20$, as $q\to \infty$ (for all $j\in [1,5]$).
In other words, the set $\{\gamma_1,\ldots, \gamma_5\}$ "tends" to the vertexes of a regular pentagon (as $q$ tends to infinity).
Someone could please give me some suggestion to deal with that?
Thanks in advance.
$f_q\to -20x^5-1$ uniformly on $|x|< 1-\epsilon$ (which contains the 5 roots of $-20x^5-1$) so for $q$ large enough, $f_q$ has $5$ roots on $|x|< 1-\epsilon$ and they converge to those of $-20x^5-1$.
For $q$ large enough $f_q$ has no roots on $|q|> 1+\epsilon$.
So (for $q$ large enough) the $q-5$ remaining roots of $f_q$ lie on the annulus $1-\epsilon\le|x|\le 1+\epsilon$.
Since $\epsilon$ is arbitrary it means that the absolute value of the remaining roots $\to 1$.
Obviously you can replace $f_q$ by $x^q+g$ with $g$ any polynomial whose roots lie on the unit disk, the result will be the same.