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Consider the given line integral below:

$$I=\oint_C \frac{dz}{z-3}$$

Here, $C$ is the circle $|z-2|=5$

The question is:

  1. Determine whether $\displaystyle\oint_C \dfrac{dz}{z-3}=0$
  2. Does your answer to Number 1. Contradict Cauchy's (Cauchy-Goursat) Theorem?

Actually i'm on chapter Complex Line Integral, so i can't use Cauchy's Integral Formula or Residual Theorem in my work bcz they haven't yet been introduced. Permissible thing is using Cauchy-Goursat Theorem that is

if $f(z)$ is analytic in a region $R$ and on its boundary $C$. Then $$\oint_C f(z)\,\mathbb dz=0$$

My attemp :

Actually i don't have an idea how to integrate it with parametrization. The last time i integrated complex line integral in another question, my answer was said to be wrong bcz of i didn't consider the branch point or something.

At least i'm trying, here is my work:

If $|z-2|=5\Rightarrow z=2+5e^{i\theta}$, then $\mathbb dz=5ie^{i\theta}\,\mathbb d \theta$

$$\begin{align} I&=\oint_C \frac{dz}{z-3}\\ &=\int_0^{2\pi} \frac{5ie^{i\theta}}{5e^{i\theta}-1}\,\mathbb d \theta\\ &=\int_0^{2\pi} \frac{i}{5e^{i\theta}-1}\,\mathbb d\theta+\int_0^{2\pi} i\,\mathbb d\theta \tag{By Long Divison}\\ &=\int_5^5 \frac{\mathbb du}{(u-1)u}+\int_0^{2\pi} i\,\mathbb d\theta \tag{By U-Substitution}\\ &=0+2\pi i \tag{i'm not sure here}\\ &=2\pi i \end{align} $$

The reason i'm using long division is to avoid $0$, bcz when i directly using u-substitution, whatever the integrand is, i have an inteval $5\leq u\leq 5$ which is the result will gives me $0$ isn't it?

Then i check on my Solution Manual Book, the answer is $0$ (it's strange), bcz the function has singularity and the singularity causes it fails to be analytic in a certain point.

What is actually the answer for my problem. And even if I tried to deviate the rules and use The Residual Theorem:

$$\underset{z=3}{\operatorname{Res}}\,\frac{1}{z-3}=1$$

$1$ is in $C$, Then i have

$$\displaystyle\oint_C \dfrac{dz}{z-3}=2\pi i$$

In short, my answer is contradict $$\displaystyle\oint_C \dfrac{dz}{z-3}=0$$ Besides, my solution manual gives me the line integral is $0$. Which one is true? Is my Solution Manual Book wrong? Or mine? Where is my mistakes? Is my answer contradict with Cauchy's Theorem? Why?

Please give me the best explanation. Especially, why does if i'm not using long division, the line integral is $0$, but with long division it gives me $2\pi i$?

Many thanks for you if you wanna help me.

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    $\begingroup$ The correct answer is $2\pi i$. This does not contradict the Gauchy-Goursat theorem because the integrand $\frac{1}{z-3}$ is not analytic in the region under consideration, namely the disk $D_5(2)$. Indeed, it has a (simple) pole at $z=3$. $\endgroup$ – Reveillark Nov 17 '19 at 2:20
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    $\begingroup$ The integral is $2\pi i$ I don't see how they get $0$. $\endgroup$ – saulspatz Nov 17 '19 at 2:22
  • $\begingroup$ Actually, could you explain to me how to solve this using "parameterization only"? Without any theorem, i can't really understand $\endgroup$ – user516076 Nov 17 '19 at 2:22
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    $\begingroup$ What's the name of the book? $\endgroup$ – sav Nov 17 '19 at 2:35
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    $\begingroup$ This one books.google.com.au/… Which chapter/question number? $\endgroup$ – sav Nov 17 '19 at 2:39
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You can use Cauchy Goursat in the region $R=\{3\}^c$ to show that $\int_C {dz \over z-3} = \int_D{dz \over z-3} $, where $D$ is the curve $t \mapsto 3+e^{it}$, and it is straightforward to evaluate this integral and see that it is non zero.

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  • $\begingroup$ One quick question, why do we have to change the region? Is that mean, Cauchy Goursat just works on curve that match with the curve? In this case, the function $z=3+e^{i\theta}$ and we have to change the region into $R=\{3\}^c$ as you said? $\endgroup$ – user516076 Nov 17 '19 at 4:45
  • $\begingroup$ You can use Cauchy Goursat to show that the two integrals have the same value, and the second one is simpler to evaluate. $\endgroup$ – copper.hat Nov 17 '19 at 5:16

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