how many roots does $p(z) = z^{10} + 100z + 1$ has in $\{z:|z|<1\}$
Can I use Rouche theorem and say that on that region $|100z|=100>2=1+1=|z^{10}|+1$
and thus the number of roots for p(z) is the same as the number of roots for $100z$ which is $1$ (the root is $0$)? is it that simple?
$p(z)=z^{10}+100z+1$ has only two real roots. This can be shown by differentiating: $p'(z)=10z^9+100$. So the only real root of $p'(z)$ is $p'(-\sqrt[9]{10})=0$. Studying the sign of this polynomial it's easy to demonstrate that $p(z)$ is decreasing from $(-\infty ; -\sqrt[9]{10})$ and increasing from $(-\sqrt[9]{10} ; +\infty$. Because $p(z)$ tends to positive infinity for $z\to \pm \infty$ and $p(-\sqrt[9]{10})<0$, $p$ has only two real roots: one between $(-\infty ; -\sqrt[9]{10}<-1)$ and the other between $(-\sqrt[9]{10} ; 0 (p(0)=1>0))$. So there is just real root of $p(z)$ whom absolute value is less than one and it isn't zero ($p(0)=1=\neq 0).