let $f(z)=z^9+z^5+8z^3+2z+1$, well, $|f(z)-z^9|=|z^5+8z^3+2z+1|<|2|^9\text{ at } |z|=2$ so $f$ has $9$ zeroes inside $|z|<2$
and $|f(z)-8z^3|=|z^9+z^5+2z+1|<5<8|z|^3\text{ at } |z|=1$ so inside $|z|<1$ $f$ has 3 zeroes, so $6$ zeroes are there inside $1