<s>[INST] Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$. 
Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.


My solution:
Set $f(z)=3$.
For every $0<\delta<1$ and $|z|=\delta$, we have $|P(z)-f(z)|\le \delta^5+2\delta^3<3\delta<3=|f(z)|$. Hence $P(z)$ and $f(z)=3$ has the same number of zeros, which is $0$, inside the open unit disk.
Now consider the behavior of $P(z)$ when $|z|=1$. If we have $P(z)=0$, then $z^5=-2z^3-3$, and $1=|z^5|=|-2z^3-3|$ when $|z|=1$ which forces $-2z^3=2$ and $z^3=-1$.
Clearly, $z=-1$ is a zero of $P(z)$.
For the other two roots of $z^3=-1$, note that: $$ P(z)=z^2\cdot z^3+2z^3+3=z^2-2+3=z^2+1=0\implies z=\pm 1 $$
Therefore we conclude that there is only one root, $z=-1$, of $P(z)$ in the closed unit disk.

However, I am asking for other solutions, such as applying the symmetric version of Rouché''s theorem like in this post. Thank you.
 [/INST] Let $z^5+2z^3+3=0$ with $|z|\le 1. $  Then $3=|-3|=|z^5+2z^3|=|z|^3\cdot |z^2+2|\le |z^2+2|\le |z|^2+|2|=|z|^2+2\le 3.$
So $|z|=1$. So let $z=e^{it}$ with $t\in \Bbb R.$ Then $-3=Re (e^{5it}+2e^{3it})=\cos 5t+2\cos 3t.$
Since $\cos 5t\ge -1$ and $2\cos 3t\ge -2,$ therefore $\cos 5t=\cos 3t=-1$. This implies $z^5=z^3=-1,$ hence $z^2=1,$ hence $z=\pm 1.$
So with $P$ standing for $(\,|z|\le 1\land z^5+2z^3+3=0\,)$ we have $$P\implies (z\in \{-1,1\}\land z^5+2z^3+3=0)\implies (z=-1)\implies P.$$
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