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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 $|z|=1$.

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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 $|z|\neq 1$.

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  • $\begingroup$ I am not sure I get the hint. $\endgroup$ – 2132123 Jul 11 '20 at 8:36
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    $\begingroup$ If $|z|=1$, then $|10z^2|=10$. What about the other side of the equation? $\endgroup$ – Pythagoras Jul 11 '20 at 9:11
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    $\begingroup$ 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! $\endgroup$ – 2132123 Jul 11 '20 at 9:21

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