To solve the given complex integral

$3\oint \frac{1}{z-3} \, dz$

where $C$ is the contour $|z|=2$ with $0 \leq \text{Arg}(z) \leq 2\pi$, we can use the residue theorem.

Step-by-Step Solution:

1. Identify the singularity: The integrand $\frac{1}{z-3}$ has a singularity at $z = 3$.

2. Determine if the singularity is inside the contour: The contour $C$ is given by $|z| = 2$. The point $z = 3$ is outside this contour because $|3| = 3$, which is greater than 2.

3. Apply the residue theorem: The residue theorem states that for a function $f(z)$ analytic inside and on a simple closed contour $C$, except for isolated singularities inside $C$,

   $\oint_C f(z) \, dz = 2\pi i \sum \text{Residues of } f(z) \text{ inside } C.$

   Since the singularity at $z = 3$ is outside the contour, there are no residues of $\frac{1}{z-3}$ inside the contour.

4. Evaluate the integral: Given that there are no singularities inside the contour $C$,

   $\oint_C \frac{1}{z-3} \, dz = 0.$

   Therefore,

   $3 \oint \frac{1}{z-3} \, dz = 3 \cdot 0 = 0.$

So, the value of the integral is $\boxed{0}$.

Do you want more details or have any questions?

Here are 5 questions you might ask next:

1. Can you explain the residue theorem in more detail?
2. How do you determine if a point is inside a contour in the complex plane?
3. What is a singularity in a complex function?
4. Can you provide examples of other complex integrals?
5. How do you handle integrals with singularities on the contour itself?

Tip: When solving complex integrals, always check if the singularities of the integrand lie inside the given contour before applying the residue theorem.