let $h(z)=z^5+\frac{1}{3}z^3+\frac{1}{4}z^2+\frac{1}{3}$ the numbers of zeros on $M$ should be determined with $M=\{ z \in \mathbb{C}: |z| < \frac{1}{2}\}$.
So I tried to apply Rouche's theorem: with $g(z)=z^5+\frac{1}{3}z^3$ amd $f(z)=\frac{1}{4}z^2+\frac{1}{3}$, I checked on the boundary $|z|=1/2$, so $|g(z)|=|z^5+\frac{1}{3}z^3| \leq \frac{7}{96} \leq |f(z)| \leq \frac{38}{96}$
so can I deduct from this that $f+g (:=h)$ and $f$ must have the same zeros in $M=\{ z \in \mathbb{C}: |z| < \frac{1}{2}\}$? and that would reduce the problem to solving
$\frac{1}{4}z^2+\frac{1}{3}=0$? (the only solutions here would not fulfill $|z| \leq 1/2$) so can I conclude that $h$ would not be having any zeros on $|z| \leq 1/2$?