The problem is asking to compute the contour integral

$I = \int_C \frac{e^{1/z}}{(3z - 4)^2} \, dz,$

where $C$ is the unit circle centered on $z = 2$. Let's proceed with solving it step-by-step:

1. Residue Theorem:

The Residue Theorem states that for a closed contour $C$ and a meromorphic function $f(z)$, the contour integral is:

$\int_C f(z) \, dz = 2\pi i \cdot \text{Res}(f(z), z_0),$

where $z_0$ is the singularity inside $C$.

2. Analyzing the Given Integral:

Function to Integrate:

$f(z) = \frac{e^{1/z}}{(3z - 4)^2}.$

• Singularity Analysis: The term $(3z - 4)^2$ introduces a pole at $z = \frac{4}{3}$ (a double pole).
  The contour $C$ is a unit circle centered at $z = 2$. Since $z = \frac{4}{3}$ is outside this circle, no singularities lie inside $C$.

3. Conclusion:

If there are no singularities inside the contour $C$, the function $f(z)$ is analytic inside $C$, and therefore, by Cauchy's Integral Theorem:

$I = \int_C f(z) \, dz = 0.$

Final Answer:

$I = 0.$

If you have additional questions or would like me to elaborate on any step, feel free to ask!

Follow-Up Questions:

1. What is the significance of the Residue Theorem in contour integration?
2. How does the order of a pole (simple, double, etc.) affect residue computation?
3. Why is $z = \frac{4}{3}$ not inside the unit circle centered at $z = 2$?
4. Can the integral be nonzero if the singularity lies exactly on the contour $C$?
5. How would the answer change if the contour $C$ were a circle centered at the origin?

Tip:

Always sketch the contour and the location of singularities in the complex plane to verify whether the singularities are inside, outside, or on the contour.