How to compute $\int_C \frac{z^4}{z^5-z-1}$ where $C$ is the radius $2$ circle centred at $(0,0)$? Apparently, because all the finite poles are within the circle, it is good enough to use the residue at infinity. But I'm not sure how to even show all the roots are bounded in norm by 2.
 A: Step1: all the roots of $z^5-z-1$ are simple and lie inside the region $\left|z\right|\leq\frac{3}{2}$.
The roots are simple because $\gcd(z^5-z-1,5z^4-1)=1$, and they lie in the region $\left|z\right|\leq\frac{3}{2}$ because if $\left|z\right|=C>\frac{3}{2}$, $\left|z\right|^5 > |z|+1 \geq |z+1|$ contradicts $z^5=z+1$.
Step2: by Cauchy's theorem, the given integral equals $2\pi i$ times the sum of the residues of the function $f(z)=\frac{z^4}{z^5-z-1}$, that also equals $\oint_{|z|=R}f(z)\,dz$ for any $R>\frac{3}{2}$. However:
$$ \oint_{|z|=R} f(z)\,dz - \oint_{|z|=R}\frac{dz}{z} = o(1) $$
for $R\to +\infty$, hence the given integral simply equals $\color{red}{2\pi i}$. This happen because for any $z$ far from the origin $f(z)-\frac{1}{z}=O\left(\frac{1}{|z|^5}\right)$, hence by integrating $f(z)-\frac{1}{z}$ over the circle $|z|=R$ we get something that is $o(1)$ for $R\to +\infty$.
A: Use Rouche's Theorem: on $\;|z|=2\;$ , we have that $\;|z+1|<|z^5|\;$ , so both $\;z^5\;$ and $\;z^5-z-1\;$ have the same number of roots in $\;|z|<2\;$ .
Define now $\;f(z):=\cfrac{z^4}{z^5-z-1}\;$ , and let now $\;z_k\;,\;\;k=1,2,3,4,5\;$ ,  be a root of our quintic, then as all these roots are simple we get that
$$Res_{z=z_k}(f)=\lim_{z\to z_k}(z-z_k)f(z)=\lim_{z\to z_k}\frac{z^4}{5z^4-1}=\frac{z_k^4}{5z_k^4-1}$$
Before going into this, you may want to observe that the roots of $\;z^5-z-1\;$ are: one real root and two pairs of complex non-real conjugate roots.
A: Let$$g(z)=\frac{5z^4-1}{z^5-z-1}\text{, let }h(z)=\frac1{z^5-z-1}.$$ and let $r$ be the greatest absolute value of the roots of $z^5-z-1$. Then $f=\frac{g+h}5$ and therefore $\int_Cf=\frac15\bigl(\int_Cg+\int_Ch\bigr)$. By the argument principle, and because $r<2$, $\int_Cg=5\times2\pi i=10\pi i$. On the other hand, if $R>r$, then the integral of $h$ along the circle of radius $R$ around $0$ does not depend on $R$. Furthermore, its limit when $R\to\infty$ is equal to $0$. Therefore$$\int_Cf=\frac{\int_Cg+\int_Ch}5=\frac{10\pi i+0}5=\color{red}{2\pi i}.$$
