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To solve this, we aim to calculate the **sum of the residues** of \( f(z) \) at all its finite singular points.

### Given function:
\[
f(z) = \frac{z^2}{(z^2 + 2)^3 (z^2 - 1)^4}
\]

### Steps:

#### 1. Identify singularities:
The singularities of \( f(z) \) occur at the roots of \( (z^2 + 2) \) and \( (z^2 - 1) \):
- Roots of \( z^2 + 2 = 0 \): \( z = \pm i\sqrt{2} \)
- Roots of \( z^2 - 1 = 0 \): \( z = \pm 1 \)

Thus, the singular points are \( z = \pm i\sqrt{2} \) and \( z = \pm 1 \).

#### 2. Nature of singularities:
- \( z = \pm i\sqrt{2} \): These are **poles of order 3** due to the factor \( (z^2 + 2)^3 \).
- \( z = \pm 1 \): These are **poles of order 4** due to the factor \( (z^2 - 1)^4 \).

#### 3. Sum of residues at finite singular points:
Using the **residue theorem**, the sum of the residues of \( f(z) \) at all finite singularities is related to the residue at infinity:
\[
\text{Sum of residues at finite points} = -\text{Residue at infinity}.
\]

#### 4. Residue at infinity:
To compute the residue at infinity, we analyze the behavior of \( f(z) \) as \( z \to \infty \). Rewrite \( f(z) \) in terms of \( w = \frac{1}{z} \) (where \( w \to 0 \) as \( z \to \infty \)):

Substitute \( z = \frac{1}{w} \), so:
\[
f(z) = f\left(\frac{1}{w}\right) = \frac{\left(\frac{1}{w}\right)^2}{\left(\left(\frac{1}{w}\right)^2 + 2\right)^3 \left(\left(\frac{1}{w}\right)^2 - 1\right)^4}.
\]

Simplify:
\[
f\left(\frac{1}{w}\right) = \frac{\frac{1}{w^2}}{\left(\frac{1}{w^2} + 2\right)^3 \left(\frac{1}{w^2} - 1\right)^4}.
\]

Multiply numerator and denominator by \( w^{18} \) (to clear all denominators):
\[
f\left(\frac{1}{w}\right) = \frac{w^{16}}{\left(1 + 2w^2\right)^3 \left(1 - w^2\right)^4}.
\]

As \( w \to 0 \), the leading term of the numerator and denominator gives the residue at infinity. Expanding the dominant term, we observe that:
\[
\text{Residue at infinity} = 0.
\]

#### 5. Conclusion:
Since the residue at infinity is zero, the sum of the residues at all finite singular points is also zero:
\[
\boxed{0}
\]

Let me know if you need further clarification or additional steps explained! Here are related questions:

1. What is the residue at \( z = 1 \)?
2. How does the order of the pole affect the residue calculation?
3. Can the residue at infinity ever be non-zero for rational functions?
4. How can we expand \( f(z) \) for large \( z \) to verify the behavior at infinity?
5. Why does the residue theorem simplify calculations for rational functions?

**Tip**: Always check the nature of singularities before computing residues to simplify your calculations.

设 f(z) = \frac{z^2}{(z^2 + 2)^3 (z^2 - 1)^4}，则 f(z) 的所有有限奇点的留数和等于（）。

Explore the calculation of the sum of residues for f(z) = z^2/[(z^2+2)^3(z^2-1)^4]. This problem leverages the residue theorem and involves poles of higher orders, rational functions, and behavior at infinity. Suitable for undergraduate mathematics students.

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