We are asked to evaluate the contour integral

$\oint_C \frac{z^2 + 1}{z} dz$

where $C$ is the circle of radius 2 centered at the origin, oriented in the positive (counterclockwise) direction.

Solution:

The given integral can be split into two simpler integrals by rewriting the integrand as follows:

$\oint_C \frac{z^2 + 1}{z} dz = \oint_C \left( z + \frac{1}{z} \right) dz$

Now, evaluate each part separately:

1. First term: $\oint_C z \, dz$

   Since $z$ is analytic (holomorphic) everywhere within and on the contour $C$, and by Cauchy's Integral Theorem, the integral of any analytic function over a closed contour is zero. Therefore:

   $\oint_C z \, dz = 0$

2. Second term: $\oint_C \frac{1}{z} \, dz$

   This is a standard contour integral known as the integral of $\frac{1}{z}$ around a contour enclosing the origin. By the residue theorem or simply from the known result, this integral equals $2\pi i$ (since the residue of $\frac{1}{z}$ at $z = 0$ is 1):

   $\oint_C \frac{1}{z} \, dz = 2\pi i$

Combining the results:

$\oint_C \frac{z^2 + 1}{z} dz = 0 + 2\pi i = 2\pi i$

Thus, the value of the contour integral is $\boxed{2\pi i}$.

Let me know if you need further clarification or more details. Here are 5 related questions to deepen your understanding:

1. What is the Cauchy's Integral Theorem, and how does it apply to analytic functions?
2. How is the residue theorem used to evaluate contour integrals?
3. Why is the integral of $\frac{1}{z}$ around a circle enclosing the origin equal to $2\pi i$?
4. Can this result be generalized to other contours enclosing singularities?
5. What changes if the radius of the circle is increased or decreased?

Tip: When solving contour integrals, always check if the integrand has singularities inside the contour to determine the best approach, such as applying the residue theorem or using Cauchy's Integral Formula.