The problem asks for the residue of the function

$g(z) = \frac{1}{(z^2 + 1)^3}$

at $z = i$.

Step 1: Simplify $g(z)$

The function can be rewritten as:

$g(z) = \frac{1}{(z - i)^3 (z + i)^3}$

since $z^2 + 1 = (z - i)(z + i)$.

Step 2: Determine the nature of the singularity at $z = i$

At $z = i$, the factor $(z - i)^3$ in the denominator causes a pole of order 3. Thus, we need to calculate the residue of this pole of order 3.

Step 3: Residue for a pole of order 3

For a pole of order $n$ at $z = i$, the residue is calculated using the following formula:

$\text{Residue} = \frac{1}{(n-1)!} \lim_{z \to i} \left[ \frac{d^{n-1}}{dz^{n-1}} \left( (z - i)^n g(z) \right) \right]$

Since the pole is of order 3, we need to find the second derivative (i.e., $n-1 = 2$) of $(z - i)^3 g(z)$, and then evaluate it at $z = i$.

Step 4: Expression for the second derivative

Let $f(z) = (z - i)^3 g(z) = \frac{1}{(z + i)^3}$. Now we find the second derivative of $f(z)$ at $z = i$.

$f'(z) = -3 (z + i)^{-4}$ $f''(z) = 12 (z + i)^{-5}$

Now, evaluate $f''(z)$ at $z = i$:

$f''(i) = 12 (2i)^{-5} = 12 \times \frac{1}{(32i^5)} = 12 \times \frac{-i}{32} = -\frac{12i}{32} = -\frac{3i}{8}$

Step 5: Conclusion

The residue of $g(z)$ at $z = i$ is $\boxed{-\frac{3i}{8}}$, which corresponds to option D.

Would you like more details or further clarifications? Here are 5 related questions to explore:

1. How do you find residues for higher-order poles?
2. What is the significance of a residue in complex analysis?
3. Can you explain the difference between simple poles and higher-order poles?
4. How is the residue theorem applied in evaluating contour integrals?
5. How can you generalize this method to functions with different types of singularities?

Tip: When calculating residues for poles of higher order, using the appropriate derivative formula is crucial. Double-check your work, especially when handling derivatives!