Confused about Rouche's theorem I need to find the zeros of $z^6-4z^5+z^2-1$ in the unit disk using Rouche's theorem. 
The hint given is to consider the functions $g(z)=z^6-4z^5+z^2-1$ and $f(z)=-4z^5$. The issue is that the inequality $|f(z)-g(z)|<|f(z)$, that is, $|z^6+z^2-1|<4|z^5|$ does not seem to hold on the unit disk. What am I doing wrong?
Also, in general, how do you find the function $g$ required to apply Rouche's theorem?

You have to check the inequality holds on the boundary. This is easy for a disk: $\lvert z \rvert =1$, so you have
$$ \lvert z^6+z^2-1 \rvert < \lvert z^6 \rvert + \lvert z^2 \rvert + 1 = 3, $$
and
$$ \lvert 4z^5 \rvert = 4>3. $$
Hence $ -4z^5 $ is larger in modulus than $z^6+z^2-1$ on the boundary of the unit disc, so $-4z^5$ and $(z^6+z^2-1)+(-4z^5)$ have the same number of roots inside.
In general, you can't tell. Here you look at the term with the largest coefficient, but there's no general way to tell; worse, it may not be possible to find such a splitting.
