Given the function: $$ f(z)=z^9 +10z^2+1 $$ The number of zeros in $1 \leq |z| \leq 2$ is found by applying Rouche's theorem: $$ Z(f) = (Z(f) \text{ in }|z| \leq 2) - (Z(f) \text{ in }|z| \leq 1) $$ On $|z|=1, |z^9| < |10z^2+1| \rightarrow Z(f) = 9$ in $|z| \leq 1$
On $|z|=2, |z^9| > |10z^2+1| \rightarrow Z(f) = 2$ in $|z| \leq 2$
Thus, $Z(f) = 9-2 = 7$ on $1 \leq |z| \leq 2$
My question is: When counting the number of zeros of $z^9$ inside $|z| \leq 2$, how do we know that all the 9 zeros of $z^9$ are inside the disk $|z| \leq 2?$