Zeroes of $2z^5-15z^2+z+2$ In preparation for qualifying exams I am working through old exams and came across the following question: 

Determine the number of roots, counted with multiplicity, of the equation $$2z^5-15z^2+z+2$$ inside the annulus $1\leq |z|\leq 2$.

It seemed like a relatively straight forward application of Rouche's Theorem and was able to show there are two roots inside the unit disk, but when I was considering the boundary of $D(0,2)$, I couldn't seem to get a strict inequality in order to apply Rouche's Theorem. For example I chose $$f(z)=-15z^2+z+2$$ and $$g(z)=2z^5$$ but the best I could do was $|f(z)|\leq 64 =|g(z)|$ on $\partial D(0,2)$. Similar problems happened on different choices for $f$ and $g$.
P.S. I am trying to use: If $|f|<|g|$ on $\partial D(0,r)$ then $|Z_{D(0,r)}(g-f)|=|Z_{D(0,r)}g|$
 A: Rouche's theorem is typically about the interior of a region (although the conditions on the boundary prevent roots there too).  In your case, you're interested in $|z|\leq 2$.  So, let's take your splitting and use $|z|=2+\varepsilon$ for some $\varepsilon>0$.  In this case,
$$
|f(z)|\leq 15(2+\varepsilon)^2+(2+\varepsilon)+2=64+61\varepsilon+\varepsilon^2
$$
and
$$
|g(z)|=2(2+\varepsilon)^5=64+160\varepsilon+160\varepsilon^2+80\varepsilon^3+20\varepsilon^4+2\varepsilon^5.
$$
Therefore, on the circle of radius $2+\varepsilon$, $|f(z)|<|g(z)|$, so there are the same number of roots of both $f$ and $g$ in the open disk $B(0,2+\varepsilon)$.  But, this is true fore every epsilon, so there can be no roots of $f$ outside the disk of radius $2$ (details hidden below).

 If there were a root of $f$ outside the disk of radius $2$, then let $r$ be the radius of this root.  Consider two radii $2+\varepsilon_1<r<2+\varepsilon_2$.  Applying Rouche's theorem to both of these values gives that $f$ has the same number of roots within disks of radius $2+\varepsilon_1$ and $2+\varepsilon_2$.  This, however, is impossible since $f$ has a least one fewer root within the smaller disk whereas $g$ does not.

For the disk of radius $1$, the $-15z^2$ should be a direct application of Rouche's theorem.
A: Actually your idea does work: With $f(z)=-15z^2+z+2$ and $g(z)=2z^5$ you have for $|z| = 2$
$$
 |f(z)| = |-15z^2+z+2| < |-15 z^2| + |z|+ 2 = 64 = |g(z)|.
$$
Here we have strict inequality in the triangle inequality because the numbers $-15z^2$, $z$ and $2$ cannot all have the same argument.
So Rouche's theorem can be applied to conclude that $f+g$ and $g$ have the same number of zeros in $D(0, 2)$.
