How many solutions of $3z^5 + z^2 + 1=0$ have in $1<|z|<2$. I used Rouche's theorem. I got $5$ solutions in $1<|z|<2$. Is my approach correct? 
 A: All the solutions $z\in\mathbb{C}$ of $3z^5+z^2+1=0$ satisfy $|z|<1$.  I am offering an elementary proof.  You can also use Rouché's Theorem to prove this.

 If $3z^5+z^2+1=0$, then  $$3\,|z|^5=\left|3z^5\right|=\left|-z^2-1\right|\leq \left|-z^2\right|+|-1|=|z|^2+1\,.$$ If $|z|\geq 1$, then we have that $$3\,|z|^5>2\,|z|^5=|z|^5+|z|^5\geq |z|^2+1\,,$$ which is a contradiction.  Thus, all the solutions $z\in\mathbb{C}$ of $3z^5+z^2+1=0$ satisfy $|z|<1$. One can improve the bound by showing that the roots must satisfy $$0.7=\frac{7}{10}<|z|<\dfrac{\sqrt[3]{20}}{3}<0.905\,.$$  It is quite a delight to see that the upper bound is very sharp.  The root with the maximum modulus has the modulus of around $0.9047$, and $\dfrac{\sqrt[3]{20}}{3}\approx 0.9048$.  The lower bound is also not bad, as the root with the minimum modulus has the modulus of roughly $0.7208$.

A: Note that for $|z|=1$, 
$$
|3z^5| = 3 > 1+1 \geq |z^2 + 1|
$$
so Rouche implies there are 5 roots inside the unit disk.  The main issue with your argument is that you need to check which function dominates on the boundary of the chosen contour.
A: An illustrative plot

As $3z^5+z^2+1 = \phi(x,y) + i \psi(x,y) = 0$ we have $\phi(x,y)=0$ in red and $\psi(x,y) = 0$ in blue.
