Finding closed paths $\gamma(a,r)$ such $\displaystyle \int_{\gamma(a,r)} \frac{5z^2-8}{z^3-2z^2}$ takes value $-2i\pi$ or $18i\pi$?
From this question it is already know that $\displaystyle \int_{\gamma(0,1)} \frac{5z^2-8}{z^3-2z^2}=4\pi i \,$ [corrected].
I'm getting very frustrated at this... I've tried using all variations of Cauchy's Theorem I know but nothing comes up!
Everything is in the question, can you help me?
On
Note: Git Gud's answer has the right idea, but unfortunately it quotes a result from the original version of the question which was incorrect, and as a result, misses a key step which the actual solution requires. This answer simply seeks to rectify that mistake, but makes references to ideas explained more fully in Git Gud's answer.
Firstly, we are given the (correct) integral, $\,\displaystyle \int_{\gamma(0,1)} \frac{5z^²-8}{z^3-2z^2}\, dz=4\pi i$, where $\gamma(0,1)\colon [0,2\pi ]\to \mathbb C, \,\,\theta \mapsto e^{i\theta}$.
What we would ideally want to do is play around with integrating around multiple loops of the pole at $z=0$ to get integrals that are integer multiples of $4\pi$. Unfortunately, this will not get you an integral which equals $-2\pi i$ or $18\pi i$.
However, the integrand also has a pole at $z=2$, so we decide to look the integral on a circular path about this. This can be done in exactly the same way in which the above integral is handled. Since this is taken for granted by the OP, I will leave this calculation as a footnote$^\dagger$; what we find is that,
$$\,\displaystyle \int_{\gamma(2,1)} \frac{5z^²-8}{z^3-2z^2}\, dz=6\pi i$$
where $\gamma(2,1)\colon [0,2\pi ]\to \mathbb C, \,\,\theta \mapsto 1 - e^{i\theta} \quad$(i.e. anticlockwise circle, centre 2, radius 1).
Note that the paths $\gamma(0,1)$ and $\gamma(2,1)$ have the point $1$ in the complex plane in common; conveniently, this is the start/end point for both paths (though this was not truly essential) so that we can legitimately just create a new path by composing the two paths together (see Git Gud's answer for more details).
Now we see that, since $-2\pi i = 4\pi i + (-6\pi i)$, we can define a new path, $$\gamma = \gamma(0,1) \star \gamma^-(2,1)$$ where $\star$ means composition of paths and the $[path]^-$ means traversing the path in the opposite direction$^\ddagger$. Then,
\begin{align} &\, \int_{\gamma} \frac{5z^²-8}{z^3-2z^2}\, dz\\ = &\,\int_{\gamma(0,1)} \frac{5z^²-8}{z^3-2z^2}\, dz\quad + \quad\int_{\gamma^-(2,1)} \frac{5z^²-8}{z^3-2z^2}\, dz\\ = &\,\int_{\gamma(0,1)} \frac{5z^²-8}{z^3-2z^2}\, dz\quad - \quad\int_{\gamma(2,1)} \frac{5z^²-8}{z^3-2z^2}\, dz\\[2.5mm] = &\quad 4\pi i - 6\pi i\\[2.5mm] = &\quad -2\pi i\\ \end{align}
Similarly, if we define the path $$\beta = 3 \times \gamma(2,1)$$ (by which we mean $\gamma(2,1)$ composed with itself 3 times), then we get,
\begin{align} &\, \int_{\beta} \frac{5z^²-8}{z^3-2z^2}\, dz\\ = \,&\, 3\,\int_{\gamma(2,1)} \frac{5z^²-8}{z^3-2z^2}\, dz\\[2.5mm] = \,&\, 3 \cdot 6\pi\\[2.5mm] = \,&\, 18\pi i \end{align}
As promised, the details:
$^\dagger$:
\begin{align} &\, \int_{\gamma(2,1)} \frac{5z^²-8}{z^3-2z^2}\, dz\\ = &\, \int_{\gamma(2,1)} \frac{f(z)}{z-2}\, dz\\ \end{align} where $f(z) := \frac{5z^²-8}{z^2}$, which is holomorphic on $\bar{B}(2,1)$ \begin{align} = &\, 2\pi i \cdot f(2) \quad \text{ by Cauchy's integral formula (see question linked in OP})\\ = &\, 6 \pi i \end{align}
$^\ddagger$:
For two paths, $\gamma_1:[a,b] \rightarrow \mathbb{C}$, $\gamma_2:[c,d] \rightarrow \mathbb{C}$, such that $\gamma_1(b) = \gamma_2(c)$, we define,
\begin{align} (\gamma_1 \star \gamma_2) \,&: \,[a, b+(d-c)]\\[2ex] &: \, t \rightarrow \begin{cases} \gamma_1(t), & \text{if }t \in [a,b]\\[2ex] \gamma_2(t-b), & \text{if }t \in [b, \, b+(c-d)] \end{cases} \end{align}
to be the composition of the two paths.
Also, given a path $\gamma_1$ (as above), we define the reverse of the path, $\gamma_1^-$ by,
$$\gamma_1^- \,:\, [a,b] \rightarrow \mathbb{C} \,:\, t \rightarrow \gamma(b-t)$$
This is a two part answer. First I hint at an easy way to answer the problem, afterwards I generalize to a way of solving similar problems, including this one.
You already know that $\displaystyle \int_{\gamma(0,1)} \frac{5z^²-8}{z^3-2z^2}\mathrm dz=-\dfrac{\pi}{4}i$, where $\gamma(0,1)\colon [0,2\pi ]\to \mathbb C, \theta \mapsto e^{i\theta}$.
Let us now consider $\displaystyle \int_{C} \frac{5z^²-8}{z^3-2z^2}\mathrm dz$, where $C\colon [0,6\pi]\to \mathbb C, \theta \mapsto e^{i\theta}$. Also let $C_1\colon [0,2\pi]\to \mathbb C, \theta \mapsto e^{i\theta}$, $C_2\colon [2\pi ,4\pi]\to \mathbb C, \theta \mapsto e^{i\theta}$ and $C_3\colon [4\pi i,6\pi]\to \mathbb C, \theta \mapsto e^{i\theta}$. (I just broke the three loops into three one-loop curves).
It follows that
$$\int_{C} \frac{5z^²-8}{z^3-2z^2}\mathrm dz=\int_{C_1} \frac{5z^²-8}{z^3-2z^2}\mathrm dz+\int_{C_2} \frac{5z^²-8}{z^3-2z^2}\mathrm dz+\int_{C_3} \frac{5z^²-8}{z^3-2z^2}\mathrm dz,$$
but since $\text{im}(C)=\text{im}(C_1)=\text{im}(C_2)=\text{im}(C_3)$, one gets
$$ \begin{align} \int_{C} \frac{5z^²-8}{z^3-2z^2}\mathrm dz&=\int_{C} \frac{5z^²-8}{z^3-2z^2}\mathrm dz+\int_{C} \frac{5z^²-8}{z^3-2z^2}\mathrm dz+\int_{C} \frac{5z^²-8}{z^3-2z^2}\mathrm dz\\ &=3\int_{C} \frac{5z^²-8}{z^3-2z^2}\mathrm dz\\ &=3\cdot \left(-\dfrac \pi 4i\right) \end{align}.$$
So this hints at how you can get all integer multiplies of $-\dfrac\pi 4i$ as the value of $\displaystyle \int \limits_{\mathop{\text{Appropriate}}_{\large \text{curve}}} \frac{5z^²-8}{z^3-2z^2}\mathrm dz.$ To change the sign of the values, just go around the circle in the other direction.
One (more general than usual) version of Cauchy's integral formula says the following.
Let $A$ be an open set, let $k\in \mathbb N$, let $r\in ]0,+\infty[$, let $z_0\in A$, let $\gamma\colon [0,2k\pi i]\to \mathbb C, \theta\mapsto z_0+re^{i\theta}$, let $I_\gamma (z_0)$ be the winding number of of $z_0$ with respect to $\gamma$ (note that it equals $k$ or $0$) and let $f\colon A\to \mathbb C$ be a holomorphic function (hence $C^\infty(A)$).
If $\text{im}(\gamma)\subseteq A$, then $$\forall n\in \mathbb N_0\left[\dfrac {2\pi i}{n!}I_\gamma(z_0)f^{(n)}(z_0)=\int _\gamma \dfrac{f(z)}{(z-z_0)^{n+1}}\mathrm dz\right].$$
In this notation you have $a=z_0$, $f$ will be whatever's useful after an appropriate decomposition of the given integrand (see your related previous question) function and your $\gamma (a,r)$ will be my $\gamma$ with $k=1$. What you need to do is consider different values of $k$ to get different winding numbers thus multiplying $\displaystyle \int_{\gamma(a,r)} \frac{5z^²-8}{z^3-2z^2}\mathrm dz$ by $k$.