How can one integrate $(z^3)/[(z-6)(z^5-z+6)]$ along the circular path $|z|=5$? My aim is to determine the value of $$\int_C f(z) \, dz = \int_{|z|=5} \frac{z^3}{(z-6)(z^5-z+6)}\,dz.$$
The quintic factor is troublesome. By splitting it as $(z^5)+(-z+6)$ and applying Rouché's Theorem, we see that all five roots are within our contour. By Decartes' Rule and by looking at its derivative and intercept, it has only one real root $\rho_1$, which is negative. (We can factor via this root, but I don't see the use: $(z-\rho_1)(z^4+\rho_1 z^3 +\rho_1^2 z^2+\rho_1^3 z +\rho_1^4-1)$.)
By writing $z^3=((z-6)+6)^3=(z-6)^3 +3(6)(z-6)^2 +3(36)(z-6)+216,$ we can reduce our problem to finding $$\int_C f(z)\,dz=216\int_{|z|=5}\frac{1}{(z-6)(z^5-(z-6))}\,dz.$$
I have not gotten much further. That the quintic can be written as $z^5-(z-6)$ or $z(z^4-1)+6$ occurred to me. I also thought that if I could show that the five roots are distinct, then the poles are simple ones and the formula $$
\operatorname*{Res}_{z=\rho_j}f(z)=\lim_{z\to \rho_j}(z-\rho_j)g(z)
$$
might apply, but then we'd express the answer in terms of the $\rho_j$?
I am thinking rather that a concrete number can be given for the answer, since this is a past qualifying exam question.
 A: For each $r>0$, let $C_r:[0,2\pi]\to\Bbb{C}$ be the curve $t\mapsto re^{it}$. Also, let $C_{r,\text{opp}}:[0,2\pi]\to\Bbb{C}$ be the oppositely oriented curve $t\mapsto re^{-it}$. Then by making the substitution $z=\frac{1}{\zeta}$ (which reverses orientation and inverts lengths) one has
\begin{align}
\int_{C_r}f(z)\,dz&=\int_{C_{1/r,\text{opp}}}f\left(\frac{1}{\zeta}\right)\,d\left(\frac{1}{\zeta}\right)\\
&=\int_{C_{1/r, \text{opp}}}f\left(\frac{1}{\zeta}\right)\left(-\frac{d\zeta}{\zeta^2}\right)\\
&=\int_{C_{1/r}}\frac{1}{\zeta^2}f\left(\frac{1}{\zeta}\right)\,d\zeta,
\end{align}
where in the last line, the minus sign cancels the opposite orientation.
Now, make this substitution, and then figure out where the poles lie (once again Rouche's theorem will be useful here), but this time, the integral is much easier to calculate.
A: Let $C_7$ be the circle with center zero, radius $7$. Let $$g(z)\,=\,\frac{1}{(z-6)(z^5-z+6)}.$$
We find the value of $\int_C g(z) \, dz = \int_{|z|=5} g(z)\,dz.$ Let $C_6$ be a tiny circle with center $6$, radius $1/10$.
We have a multiply-connected domain formed by circles $C,C_6,C_7$, in which $g(z)$ is analytic. Hence,
$$\int_{C_7} g(z) \, dz \,-\, \int_C g(z) \, dz \,-\, \int_{C_6} g(z) \, dz \,=\,0.$$
Since all six singular points of $g(z)$ are within $C_7,\,\,\int_{C_7} g(z) \, dz \,=\,2\pi i\, \operatorname*{Res}_{z=0}\phi(z),$ where
$$\phi(z)\,=\,\frac{1}{z^2}g\left(\frac{1}{z}\right)\,=\,\frac{z^4}{(1-6z)(1-z^4+6z^5)}.$$
Since this $\phi$ is analytic at zero, we obtain $=\int_{C_7} g(z) \, dz \,=\,0.$
Next note $t(z)\,=\,\frac{1}{z^5-z+6}$ is analytic in the disc $D(6,1/2).$ By the Cauchy Integral Formula, $$t(6)\,=\,\frac{1}{2\pi i}\int_{C_6}\frac{t(z)}{z-6}\, dz\,=\,\frac{1}{2\pi i}\int_{C_6}g(z)\, dz.$$
So finally $$\int_C f(z)\,dz=216\int_{C}g(z)\,dz\,=\,- 216\int_{C_6} g(z) \, dz\,=\,6^3(-2\pi i) \,t(6)\,=\,\frac{-\pi i}{18}.$$
