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Determine the number of zeros, with multiplicity, of the polynomial $f(z) = 1+4z^3 +z^{10} +2z^{12}$ inside the annulus $\{z \in \mathbb{C}:1<|z|<2\}$

First we consider the unit disk $|z|<1$.

Let $g(z)=-z^{10}$

Then $|f(z)+g(z)|= |1+4z^3+2z^{12}|= 7 < 8 = |f(z)|$

So by Rouche's theorem, $f $ and $g$ have the same number of zeros. Since $g$ has 10, $f$ has 10.

Now consider the disk $|z|<2$.

Let $g(z)=-2z^{12}$.

Then $|f(z)+g(z)|= |1+2^4+2^{10}| < 2^{13}$.

So So by Rouche's theorem, $f $ and $g$ have the same number of zeros. Since $g$ has 12, $f$ has 12.

Thus, in $\{z \in \mathbb{C}:1<|z|<2\}$, we get 2 zeros.

Now for the multiplicity, usually we want to check the derivative.

So we have $f(z) = 1+4z^3 +z^{10} +2z^{12}$ and $f'(z)= 12z^2+10z^9+24z^{11}$. But from here, I'm not sure how to find the zeros.

Thanks in advance!

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    $\begingroup$ $|f(z)| = 8$ $(|z|=1)$ is wrong. In fact $f(-1)=0$. $\endgroup$
    – Gerd
    Commented Aug 11, 2022 at 7:52
  • $\begingroup$ For $|z|=1$ the term $4z^3$ is dominating. So you should take $g(z)=4z^3.$ For $f'$ the term $24z^{11}$ is dominating for $|z|=2$ and for $|z|=1.$ $\endgroup$ Commented Aug 11, 2022 at 13:22
  • $\begingroup$ @RyszardSzwarc How to show that $4z^3$ is dominating? Note that the inequality of Rouche's theorem on the boundary is strict. Moreover, Rouche's theorem includes the multiplicity of the roots and therefore, I think that there's no need to check $f'(z)$. $\endgroup$
    – on1921379
    Commented Aug 11, 2022 at 14:23
  • $\begingroup$ @on1921379 My bad. I have missed one term. Concerning multiplicity you cannot rely on Rouche theorem without studying $f'.$ For example $f(z)= 2z^n-1$ has simple roots but $2z^n$ doesn't, in the open unit ball. $\endgroup$ Commented Aug 11, 2022 at 16:27
  • $\begingroup$ @RyszardSzwarc You are right, Thanks. $\endgroup$
    – on1921379
    Commented Aug 11, 2022 at 16:57

1 Answer 1

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As was proved by the asker the function $f(z)=1+4z^3+z^{10}+2z^{12}$ has $12$ roots in the open unit ball $|z|<2$ as the term $2z^{12}$ is dominating on the circle $|z|=2.$

For $0<\delta<1$ consider the circle $|z|=1-\delta.$ When $\delta\to 0^+$ we get $$ 4|z|^3= 4-12\delta+o(\delta^2)$$ $$1+|z|^{10}+2|z|^{12}= 1+(1-10\delta)+2(1-12\delta)+o(\delta^2)=4-34\delta+o(\delta^2) $$ Therefore for $\delta>0$ small enough the term $4z^3$ is dominating on the circle $|z|=1-\delta.$ Hence the function $f(z)$ has $3$ roots in the open ball $|z|<1-\delta$ for every $\delta>0$ small enough. Hence $f(z)$ has $3$ roots in $|z|<1.$

It remains to determine the number of roots on the circle $|z|=1.$ Assume $z_0$ satisfies $f(z_0)=0$ and $|z_0|=1.$ Then $$-4z_0^3=1+z_0^{10}+2z_0^{12}$$ hence $$4=|1+z_0^{10}+2z_0^{12}|$$ The equality implies $z_0^{10}=z_0^{12}=1,$ i.e. $z_0=\pm 1.$ We have $f(-1)=0$ and $f(1)\neq 0.$

Summarizing the function $f(z)$ has $12-3-1=8$ roots in the region $1<|z|<2.$

Concerning multiplicity, for every root $z_0$ with multiplicity greater than $1,$ there holds $f'(z_0)=0.$ We have $$f'(z)= 12z^2+10z^9+24z^{11} $$ The term $24z^{11}$ is dominating on the circle $|z|=1$ as well as Therefore all $11$ roots of $f'(z)$ are located in the open ball $|z|<1.$ Thus all the roots in the region $1\le |z|<2$ are single.

Remark It turns out that also $3$ roots of $f(z)$ located in $|z|<1$ are single as well. It follows from the fact that (according to Wolphram Alpha GCD algorithm or this) the polynomials $f(z)$ and $f'(z)$ are relatively prime, i.e. they do not have a common divisor. That means they do not have a common root.

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    $\begingroup$ Very nice explanation.+1 $\endgroup$ Commented Aug 13, 2022 at 1:09

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