Number of solutions of $z^5+4z^3=e^{iz}$ in $A=\{z:1 < |z| < 3\}.$

Question

So we need the number of solutions in an this annulus and we will use Rouche's theorem. Let $p(z)=z^5+4z^3-e^{iz}$. We will first find the number of zeros in $D[0,3]$ and then subtract from this the number of zeros in $D[0,1]$. Let's choose $f(z)=z^5$ since it's the dominating term and let $g(z)=4z^3-e^{iz}$, so we have that $h(z)=f(z)+g(z).$ Now both $f$ and $g$ are analytic, also

$$|g(z)|\leq 4|z|^3+|e^{iz}|\le 4\cdot 3^3+e^3<4\cdot3^3+3^3=5\cdot3^3<3^5=|f(z)|.\tag1 $$

According to Rouche's we have that $f(z)+g(z)$ has the same number of zeros as $f(z)$, and $f(z)=z^5$ so we got $5$ zeros in $D[0,3]$.

Let now instead $f(z)=4z^3$ and $g(z)=z^5-e^{iz}$. Now we have that

$$|g(z)|\leq|z|^5+|e^{iz}|\leq 1+ e < 1+3=4=4|z|^3=|f(z)|.\tag{2}$$

Here, Rouche's gives that we have $3$ zeros in $D[0,1]$ since $f(z)=4z^3.$ Thus, the total number of zeroes in $A$ is $5-3=2.$

QUESTIONS:

1. I don't really understand, in $(1)$, how we bound $|e^{iz}|$ by $e^3$. My reasoning is that $|e^{iz}|=|e^{ix}||e^{-y}|=|e^{-y}|\leq1$ for $y\geq0.$ And bounding with $1$ still keeps the inequality valid.
2. In the second part, when finding the number of zeros in $D[0,1]$ we just choose $f(z)=$ something else. What rules are there for choosing $f$ and $g$? I heard that $f$ should be the dominating term and this makes sense. But now we just don't use it here. Why?

Answer 1

1. You are right here.
2. “Dominant” is context-dependent. In, say, $z^5+z+1$, $z^5$ is dominant if $\lvert z\rvert$ is large, whereas $1$ is dominant if $\lvert z\rvert$ is small.