Finding closed paths $\gamma(a,r)$ such $\int_{\gamma(a,r)} \frac{5z^2-8}{z^3-2z^2}$ takes value $-2i\pi$ or $18i\pi$?

Question

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?

Answer 1

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.

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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.

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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 $k$.

Answer 2

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.

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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}

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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,

$$\gamma_1^- \,:\, [a,b] \rightarrow \mathbb{C} \,:\, t \rightarrow \gamma(b-t)$$