# Find the number of zeroes of $p(z)=z^6+z^3+10z^2+4z+3$ inside the annulus $1<|z|<2$

Find the number of zeroes (counting multiplicity) of $$p(z)=z^6+z^3+10z^2+4z+3$$ inside the annulus $$1<|z|<2$$.

I think this can be solved using Rouché’s theorem. First consider $$|z|\leq 1$$ on the boundary $$|z|=1$$ and $$|z^6+z^3|<|10z^2+4z+3|$$ which can be shown using triangle inequalities, the left side is at most $$2$$ while right side is at least $$3$$. We could also consider $$|z|\leq 1+ \epsilon$$ and our proof still works. Thus $$p(z)$$ has 2 roots inside $$|z|\leq 1$$. Also for $$|z|= 2$$ we have $$|z^6+z^3|>|10z^2+4z+3|$$. Thus we have 6 zeroes of $$p(z)$$ inside $$|z|<2$$ and therefore 4 of them are in $$1<|z|<2$$. Is this correct? The only "tricky" part is ruling out that there are zeros on $$|z|=1$$.

Your argument looks fine. To finish up, you need to show that $$p(z)\neq 0$$ on $$|z|=1$$. This can be seen by using the following
Hint: Use $$p(z)=0\Rightarrow 10z^2=-(z^6+z^3+4z+3),$$ which shows that $$|z|\neq 1$$.
• If $|z|=1$, then $|10z^2|=10$. What about the other side of the equation? – Pythagoras Jul 11 '20 at 9:11
• Ah, what i did is applied Rouche's to $|z|<1+\epsilon$ and saw it has the same number of zeroes as $|z|<1$ so no zeroes on $|z|=1$ but this is simpler. Thank you! – 2132123 Jul 11 '20 at 9:21