Define $f(z)=z^4-4z^3+8z-2$. Find how many zeros (including multiplicity) the function has in $\{z\in\mathbb{C}:|z|<3\}$.
I tried using Rouché's-theorem on $\{z\in\mathbb{C}:|z|<3\}$. The dominant factor is $z$, so $$|f(z)|\le|z+8z|=g(z)$$g(z) has only one root so $f(z)$ also has one root. for $1<|z|<3$$$|f(z)|<|z^4+8z|=h(z)$$ and $h(z)$ has three roots in this range (the ones for the equation $z^3=-8$) so I get it has four roots in this domain but when I solve it using MATLAB, I get that in $1<|z|<3$ there are two roots and not three. What am I doing wrong?
Take $g(z) = z^4-4 z^3$. Clearly if $|z|=3$ we have $g(z) \neq 0$.
If $|z| = 3$, then $|f(z)-g(z)| = |8 z-2| \le 26 < 27 = |z|^3 (4-3) \le |z|^3|z-4| = |g(z)|$.
Hence $f,g$ have the same zeros in $|z|<3$.