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 $\boxed{0}$.

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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!