How many complex roots? Find the number of solutions, counting multiplicity, in the domain $\{z \in \mathbb{C} : 1 < |z| <2\}$ to the equation 
$$z^9+z^5-8z^3+2z+1=0$$
Notice that along $e^{i \theta}$ where $\theta \in [0,2\pi]$ we have $|8z^3| $dominating the polynomial. Thus, by Rouche's theorem, there are 3 zeroes within that contour. Again, notice that $|z^9|$ dominates on $2e^{i \theta}$ again with $\theta \in [0, 2\pi]$ which gives 9 zeroes. Thus, the annulus contains 6 zeroes. Is this correct?
 A: The original question was how to calculate the number of roots, now the question has changed to confirming that the number the OP got was correct.
Sketch:
Use Rouche's theorem to count the number of roots within the disk $|z|<2$ and then use Rouche's theorem to count the number of roots within the disk $|z|<1$.  A reverse triangle inequality will take care of the disk $|z|=1$ to show that there's no root there.
Details:
For the disk $|z|<2$, write your polynomial as $z^9+(z^5-8z^3+2z+1)$.  Using the triangle inequality, we see on $|z|=2$, $|z^9|=512$ while $|z^5-8z^3+2z+1|\leq 32+64+4+1=101<512$, so we can apply Rouche's theorem.  In particular, $z^9+(z^5-8z^3+2z+1)$ and $z^9$ have the same number of roots in the disk $|z|<2$.  Since $z^9$ has $9$ roots, so does $z^9+(z^5-8z^3+2z+1)$.
For the disk $|z|<1$, write your polynomial as $-8z^3+(z^9+z^5+2z+1)$.  On $|z|=1$, $|-8z^3|=8$, while, by the triangle inequality, $|z^9+z^5+2z+1|\leq 1+1+2+1=5<8$.  Therefore, we can apply Rouche's theorem.  In particular, $-8z^3+(z^9+z^5+2z+1)$ and $-8z^3$ have the same number of roots in the disk $|z|<1$.  Since $-8z^3$ has $3$ roots, so does $-8z^3+(z^9+z^5+2z+1)$.
Therefore, there are $6$ roots in the annulus $1\leq |z|<2$.  Now, to deal with $|z|=1$, on this circle, $|-8z^3+(z^9+z^5+2z+1)|\geq |-8z^3|-|z^9+z^5+2z+1|\geq 8-5=3$.  Therefore, there are no roots on $|z|=1$, and so there are $6$ roots in the annulus $1<|z|<2$.
A quick check with WolframAlpha confirms this computation.
