We are tasked with evaluating the contour integral of $g(z)$ around the circle $|z - i| = 2$ for two different functions of $g(z)$. The contour is a circle of radius 2 centered at $z = i$.

To solve this, we will use the Residue Theorem, which is useful for evaluating integrals over closed contours when the integrand has isolated singularities inside the contour.

Problem (a)

For $g(z) = \frac{1}{z^2 + 4}$, we can rewrite it as: $g(z) = \frac{1}{(z - 2i)(z + 2i)}$ The function has two singularities at $z = 2i$ and $z = -2i$.

Now we check which of these singularities lie inside the contour $|z - i| = 2$, which is centered at $z = i$ with radius 2.

• $z = 2i$ is at a distance of $|2i - i| = |i| = 1$ from the center, so it lies inside the contour.
• $z = -2i$ is at a distance of $|-2i - i| = |-3i| = 3$ from the center, so it lies outside the contour.

Since only $z = 2i$ is inside the contour, we will calculate the residue of $g(z)$ at $z = 2i$.

Finding the residue at $z = 2i$:

The residue of $g(z) = \frac{1}{(z - 2i)(z + 2i)}$ at $z = 2i$ is: $\text{Res}(g(z), 2i) = \lim_{z \to 2i} (z - 2i)g(z) = \lim_{z \to 2i} \frac{1}{z + 2i} = \frac{1}{2i + 2i} = \frac{1}{4i}$

Now, applying the Residue Theorem: $\oint_{|z - i| = 2} g(z) \, dz = 2\pi i \times \text{Res}(g(z), 2i) = 2\pi i \times \frac{1}{4i} = \frac{\pi}{2}$

Thus, the value of the integral in part (a) is: $\boxed{\frac{\pi}{2}}$

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Problem (b)

For $g(z) = \frac{1}{(z^2 + 4)^2}$, we rewrite it as: $g(z) = \frac{1}{\left((z - 2i)(z + 2i)\right)^2}$ This function has second-order poles at $z = 2i$ and $z = -2i$.

As in part (a), $z = 2i$ lies inside the contour $|z - i| = 2$, and $z = -2i$ lies outside.

Thus, we need to compute the residue of $g(z)$ at $z = 2i$, but since this is a second-order pole, the calculation is different.

Residue at a second-order pole:

The residue at a second-order pole is given by: $\text{Res}(g(z), 2i) = \lim_{z \to 2i} \frac{d}{dz} \left( (z - 2i)^2 g(z) \right)$ First, simplify $(z - 2i)^2 g(z)$: $(z - 2i)^2 g(z) = \frac{(z - 2i)^2}{((z - 2i)(z + 2i))^2} = \frac{1}{(z + 2i)^2}$

Now, differentiate this with respect to $z$: $\frac{d}{dz} \left( \frac{1}{(z + 2i)^2} \right) = -\frac{2}{(z + 2i)^3}$ At $z = 2i$: $\text{Res}(g(z), 2i) = -\frac{2}{(2i + 2i)^3} = -\frac{2}{(4i)^3} = -\frac{2}{64i^3} = -\frac{2}{64(-i)} = \frac{2i}{64} = \frac{i}{32}$

Using the Residue Theorem again: $\oint_{|z - i| = 2} g(z) \, dz = 2\pi i \times \frac{i}{32} = \frac{-2\pi}{32} = -\frac{\pi}{16}$

Thus, the value of the integral in part (b) is: $\boxed{-\frac{\pi}{16}}$

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Summary of results:

(a) $\int_{|z - i| = 2} \frac{1}{z^2 + 4} \, dz = \frac{\pi}{2}$

(b) $\int_{|z - i| = 2} \frac{1}{(z^2 + 4)^2} \, dz = -\frac{\pi}{16}$

Do you want more details on these calculations or have any questions?

Here are 5 related questions:

1. How do you compute residues for higher-order poles?
2. What are the applications of the Residue Theorem in complex analysis?
3. Can you explain how to use Laurent series for residue calculations?
4. How does the choice of contour affect the integral in the Residue Theorem?
5. What happens if both singularities are inside the contour?

Tip: For higher-order poles, always differentiate the appropriate power of the function and evaluate at the pole to get the residue.