Let's go through your work to verify if it is correct.

Step 1: Identify the function and singularities

You are evaluating the integral of $g(z) = \frac{1}{z^2 + 4}$ around the circle $|z - i| = 2$.

You correctly factored $z^2 + 4 = (z - 2i)(z + 2i)$, which identifies two singular points at $z = 2i$ and $z = -2i$.

Step 2: Checking which singularity lies inside the contour

Given that the contour is $|z - i| = 2$, we know the center is $i$ and the radius is 2. The point $z = 2i$ is inside this contour, while $z = -2i$ is outside. This is consistent with what you wrote.

Step 3: Applying Cauchy Integral Formula

You applied the Cauchy Integral Formula for a function with a simple pole: $\oint_{\gamma} g(z) \, dz = 2\pi i \cdot \text{Res}(g(z), z = 2i)$

Step 4: Residue Calculation

You calculated the residue of $g(z)$ at $z = 2i$. The residue is given by: $\text{Res}(g(z), z = 2i) = \lim_{z \to 2i} \frac{1}{z + 2i} = \frac{1}{4i}$

You then correctly applied the formula: $\oint_{\gamma} g(z) \, dz = 2\pi i \cdot \frac{1}{4i} = \frac{\pi}{2}$

Conclusion:

Yes, your work is correct! The integral evaluates to $\frac{\pi}{2}$, just as you concluded.

Would you like more clarification or additional examples?

Here are 5 relative questions:

1. How would the result change if the contour included both singularities?
2. What is the residue theorem, and how does it generalize the Cauchy Integral Formula?
3. Can you derive the integral using partial fraction decomposition?
4. How would you approach the integral if the function had higher-order poles?
5. What if the contour was not a circle but an arbitrary shape enclosing the singularities?

Tip: When dealing with integrals of rational functions, always check if any of the singularities lie on the contour, as the results may not hold in such cases.