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.

$f(z)=z^5+2z^3+3 \Rightarrow f'(z)=5z^4+6z^2>0 \Rightarrow f(z)$ is an increasing function. So $f(z)=0$ can have at most oone real root and it is $z=-1.$
Take $$-3=z^5+2z^3 \Rightarrow 3=|z^5+2z^3|\le |z|^5+2|z|^3 \le 3.$$
This proves that at least one root of this equation is s.t. |z|=1.
Other roots cannot not satisfy $|z|<1$, so they will lie outside the unit disk i.e.,$z>1$
In fact, one can numerically check that apart from -1, this equation has two pairs non-real complex conjugate roots roots such that $|z|>1.$
