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.

Note that on $|z| < 1$ there are no roots as $|z^5+2z^3| < 3$ and so the roots of $z^5+2z^3+3$ on $|z| \leq 1$ lie on the boundary $|z| = 1$. Let $z = cos(\theta) + isin(\theta)$ satisfy the polynomial relation $P(z)$ stated above. 
We must have $|(e^{i\theta})^5 + 2(e^{i\theta})^3|^2 = 9$  and hence $|e^{i2\theta}+2|^2 = 9$ which is exactly when $\theta \equiv 0$ $  mod(\pi)$ so there are atmost two roots of $P(z)$ in $|z| = 1$ at $z=1$ and $z=-1$; but only $z = - 1$ is a root. Your conclusion should be correct.
