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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!

Le résidu de g(z) = 1 / (z^2 + 1)^3 en z = i est?

Explore how to calculate the residue of g(z) = 1 / (z^2 + 1)^3 at z = i. This solution covers advanced concepts in complex analysis, focusing on the calculation of residues at higher-order poles using derivatives. Suitable for college-level students.

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