Confused on evaluating line integral with parameterization

Question

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.

Answer 1

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.