This is from an exam meant for high-school students:
Suppose $z$ is a complex number satisfying $z^5+z^3+z+3=0$. Then |z| satisfies: A. $|z|\lt 1$ B. $|z|\geq 1$ C. $|z|\leq \frac12$ D. $|z|\leq \frac13$
Now, the solution is pretty simple. Either you recognise that $-1$ is a solution to the equation so option B is the only feasible option, or you just notice that if $|z|\lt 1$ then $|z|^5+|z|^3+|z|\lt 3$ which is a contradiction.
However, I want to use Rouche’s Theorem because I view it as a more powerful tool than the ones that are currently provided to us and want to be able to use it if required. I do not, at all, understand the conditions required to be satisfied for the theorem to hold, but I (hopefully justifiably) assume that the conditions hold true for polynomials. I also think that in choosing f(z) and g(z), we must keep in mind that f(z) must have 5 roots in the disc $|z|\leq 1$ so that all roots of $f(x)-g(x)$ satisfy $|z|\leq 1$. So basically I need help in choosing the two polynomials f(z) and g(z).
Put $f(z)=3$ and $g(z)=z^5+z^3+z$. Then $|z|=r<1\implies\bigl|g(z)\bigr|<\bigl|f(z)\bigr|$ and therefore Rouché's theorem tells us that $f$ and $f+g$ have the same number of zeros at the open disk with center $0$ and radius $r$. But $f$ has no zero there and $f(z)+g(z)=z^5+z^3+z+3$. So, $z^5+z^3+z+3$ has no root with absolute value smaller than $1$.