text: # Rouche's Theorem application and find mulitplicity
  
  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.
  
  • $|f(z)| = 8$ $(|z|=1)$ is wrong. In fact $f(-1)=0$.
  – Gerd
  Aug 11, 2022 at 7:52
  • 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 $|z|=1.$ Aug 11, 2022 at 13:22
  • @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 $f'(z)$. Aug 11, 2022 at 14:23
  • @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. Aug 11, 2022 at 16:27
  • @RyszardSzwarc You are right, Thanks. Aug 11, 2022 at 16:57
  
  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.
  
  • Very nice explanation.+1 Aug 13, 2022 at 1:09
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