 Evaluate
 i.
 ii.
 C
 C
 sin2 (z)
 (z − π
 6
 )3 
dz where C is |z| = 2,
 e2z
 (z −i)4 
dz where C is |z| = 2

Let's evaluate both contour integrals, where \( C \) is the circle \( |z| = 2 \) in each case.

### i. \( \int_C \frac{\sin^2(z)}{(z - \frac{\pi}{6})^3} \, dz \)

This is a contour integral involving a non-analytic function at \( z = \frac{\pi}{6} \), which is a singularity of order 3. We can apply the **residue theorem** to evaluate it if this singularity lies inside the contour.

- **Residue Theorem**: For a contour \( C \) enclosing the singularities of a function \( f(z) \), the contour integral is given by:
  \[
  \int_C f(z) \, dz = 2\pi i \sum \text{Res}(f, z_k)
  \]
  where \( z_k \) are the singularities inside \( C \).

The contour \( C \) is \( |z| = 2 \), so the singularity at \( z = \frac{\pi}{6} \) must be checked to see if it lies inside the contour:
- \( \frac{\pi}{6} \approx 0.52 \), which is less than 2, so this singularity is inside the contour.

Now we compute the residue of \( \frac{\sin^2(z)}{(z - \frac{\pi}{6})^3} \) at \( z = \frac{\pi}{6} \). Since the pole is of order 3, we need to calculate the residue of this higher-order pole, which requires taking derivatives.

The residue calculation here is more involved and requires working with the Laurent series, but ultimately the integral would reduce to:
\[
\int_C \frac{\sin^2(z)}{(z - \frac{\pi}{6})^3} \, dz = 0
\]
because the integral of \( \frac{1}{(z - z_0)^n} \) for \( n \geq 2 \) over a closed contour around \( z_0 \) is zero.

### ii. \( \int_C \frac{e^{2z}}{(z - i)^4} \, dz \)

This is another contour integral with a singularity at \( z = i \), which is of order 4. Again, we can use the **residue theorem**.

The singularity is at \( z = i \). Let's check if it lies inside the contour \( |z| = 2 \):
- \( |i| = 1 \), which is less than 2, so the singularity is inside the contour.

For a pole of order 4, the residue can be calculated using the formula for higher-order poles:
\[
\text{Res}(f, z_0) = \frac{1}{(n-1)!} \lim_{z \to z_0} \frac{d^{n-1}}{dz^{n-1}} \left[ (z - z_0)^n f(z) \right]
\]
where \( n = 4 \) in this case.

However, for this integral, since the singularity is of order 4, the contour integral will be zero, because the residue calculation for poles of order higher than 1 results in a zero contribution to the integral.

Thus:
\[
\int_C \frac{e^{2z}}{(z - i)^4} \, dz = 0
\]

### Summary of Results:

i. \( \int_C \frac{\sin^2(z)}{(z - \frac{\pi}{6})^3} \, dz = 0 \)

ii. \( \int_C \frac{e^{2z}}{(z - i)^4} \, dz = 0 \)

Would you like more details on the residue calculation process or have any further questions?

Here are 5 related questions for deeper exploration:

1. How is the residue theorem applied to contour integrals?
2. What is the procedure for calculating residues for higher-order poles?
3. How do you check whether a singularity lies inside a given contour?
4. What is the role of the Laurent series in residue calculations?
5. Why do higher-order poles (n > 2) result in a zero integral in certain cases?

**Tip**: Always verify the location of singularities relative to the contour before applying the residue theorem.

Evaluate i. ∫C (sin^2(z)/(z - π/6)^3) dz where C is |z| = 2, ii. ∫C (e^(2z)/(z - i)^4) dz where C is |z| = 2

This page explores the evaluation of contour integrals involving higher-order poles. Learn how to apply the residue theorem to solve ∫C (sin^2(z)/(z - π/6)^3) dz and ∫C (e^(2z)/(z - i)^4) dz where C is |z| = 2. Suitable for university students studying complex analysis.

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