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.
Thanks in advance!
 A: 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.
