Let $p(z)=2z^4-2z^3+2z^2+7$. I'm trying to determine the number of roots of $p(z)$ that lie inside $\{z\in\mathbb{C}:|z|<2\}$
I used Rouche's theorem and I'm trying to find find $f(z)$ such that $|f(z)-g(z)|<|f(z)|$.
Let $f(z)=z^4$, $g(z)=2z^4-2z^3+2z^2+7$. Let $|z|=2$.
I get
$|f(z)-g(z)|\\=|-z^4+2z^3-2z^2-7|\\ \le|z|^4+2|z|^3+2|z|^2+|7|=47\not\le z^4=16$
But if I let $f(z)=2z^4$, $g(z)=2z^4-2z^3+2z^2+7$ then I get $|f(z)-g(z)|\\=|2z^3-2z^2-7|\\ \le2|z|^3+2|z|^2+|7|=31\le 2z^4=32.$
So is the number of roots of $p(z)$ that lie inside $\{z\in\mathbb{C}:|z|<2\}$, $4$? If so, whats the difference between $f(z)=2z^4$ and $f(z)=z^4$?