the value of complex integral of 1/((z^3)sin(z^2)) around |z-i|=3/2, counterclockwise

To solve the complex integral of the function \( \frac{1}{z^3 \sin(z^2)} \) around the contour \( |z - i| = \frac{3}{2} \) counterclockwise, we apply the residue theorem, which is useful when dealing with integrals around closed contours in the complex plane. Let's go through the solution step-by-step.

### Step 1: Contour and Function Setup
The contour is \( |z - i| = \frac{3}{2} \), which is a circle centered at \( z = i \) with radius \( \frac{3}{2} \).

The given function is:
\[
f(z) = \frac{1}{z^3 \sin(z^2)}
\]
We need to determine the singularities of this function that lie within the contour.

### Step 2: Identify Singularities
- \( z = 0 \) is a pole of order 3 due to the \( \frac{1}{z^3} \) term.
- \( \sin(z^2) = 0 \) when \( z^2 = n\pi \), i.e., \( z = \pm \sqrt{n\pi} \) for integer \( n \). These points are also singularities of the function because the sine function vanishes at those points.

#### Let's analyze which singularities are inside the contour \( |z - i| = \frac{3}{2} \):

- The singularity \( z = 0 \) is outside the contour because \( |0 - i| = 1 \), which is less than \( \frac{3}{2} \), so we need to consider it.
- The singularity of \( \sin(z^2) = 0 \) nearest to \( z = i \) occurs when \( z^2 = \pi \), i.e., \( z = \pm \sqrt{\pi} \). Numerically, \( \sqrt{\pi} \approx 1.772 \), which lies **outside** the contour.

Thus, the only singularity inside the contour is \( z = 0 \).

### Step 3: Apply the Residue Theorem
The residue theorem states that for a meromorphic function \( f(z) \), the integral around a closed contour \( C \) is given by:
\[
\oint_C f(z) \, dz = 2\pi i \sum \text{Res}(f, z_k)
\]
where \( z_k \) are the singularities inside the contour. In our case, the only singularity inside the contour is \( z = 0 \), which is a pole of order 3.

### Step 4: Compute the Residue at \( z = 0 \)
To compute the residue of \( f(z) = \frac{1}{z^3 \sin(z^2)} \) at \( z = 0 \), we expand \( \sin(z^2) \) as a power series:
\[
\sin(z^2) = z^2 - \frac{z^6}{3!} + O(z^{10})
\]
For small \( z \), we approximate \( \sin(z^2) \) by its leading term:
\[
\sin(z^2) \approx z^2
\]
Thus, near \( z = 0 \), the function \( f(z) \) behaves as:
\[
f(z) \approx \frac{1}{z^3 \cdot z^2} = \frac{1}{z^5}
\]
Now, we need the residue of \( \frac{1}{z^5} \) at \( z = 0 \), which corresponds to the coefficient of \( \frac{1}{z} \) in the Laurent series expansion. Since \( \frac{1}{z^5} \) has no \( \frac{1}{z} \) term, the residue is zero.

### Step 5: Conclusion
Since the only singularity inside the contour is \( z = 0 \), and the residue of the function at this point is zero, the value of the integral is:
\[
\oint_{|z - i| = 3/2} \frac{1}{z^3 \sin(z^2)} \, dz = 0
\]

### Final Answer:
The value of the complex integral is \( \boxed{0} \).

---

Would you like further details or clarification on any part of this solution? Here are some related questions you might find interesting:

1. What is the residue theorem, and how is it applied in complex analysis?
2. How do you compute residues for higher-order poles?
3. What are the Laurent series, and how do they relate to singularities?
4. How can we determine which singularities are inside a given contour?
5. What are the key steps to expanding functions in power series?

**Tip:** Understanding how to classify singularities (removable, poles, and essential) is crucial in complex analysis, especially when applying the residue theorem!

The value of complex integral of 1/((z^3)sin(z^2)) around |z-i|=3/2, counterclockwise

Solve the complex integral of 1/(z^3sin(z^2)) around the contour |z - i| = 3/2 using the Residue Theorem. Learn how to identify singularities and apply the Laurent series expansion. This step-by-step solution is suitable for undergraduate students studying complex analysis.

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