To evaluate the contour integral $\int_C \frac{e^{3z}}{z^2 + 1} \, dz$, where $C$ is the contour $|z| = \frac{1}{2}$, we'll use the residue theorem, which is a powerful tool in complex analysis.

Step 1: Identify the Singularities

The integrand is $\frac{e^{3z}}{z^2 + 1}$. The function $z^2 + 1$ has roots at $z = \pm i$. These are the singularities of the function. Specifically, these singularities are poles of order 1.

• The poles are at $z = i$ and $z = -i$.

Step 2: Check if the Poles are Inside the Contour $C$

The contour $C$ is $|z| = \frac{1}{2}$, which is a circle of radius $\frac{1}{2}$ centered at the origin.

• The pole at $z = i$ has $|i| = 1$.
• The pole at $z = -i$ has $|-i| = 1$.

Both poles $z = i$ and $z = -i$ are outside the contour $|z| = \frac{1}{2}$, so there are no singularities inside the contour $C$.

Step 3: Apply the Residue Theorem

The residue theorem states that the contour integral of a function around a closed contour $C$ is $2\pi i$ times the sum of the residues of the function inside $C$.

Since there are no singularities inside the contour $C$, the sum of the residues inside $C$ is zero. Thus:

$\int_C \frac{e^{3z}}{z^2 + 1} \, dz = 0$

Conclusion

The value of the contour integral $\int_C \frac{e^{3z}}{z^2 + 1} \, dz$, where $C$ is the contour $|z| = \frac{1}{2}$, is 0.

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Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. How would the answer change if the contour $C$ was $|z| = 2$?
2. What is the residue of $\frac{e^{3z}}{z^2 + 1}$ at $z = i$?
3. How does the residue theorem apply to higher-order poles?
4. Can the residue theorem be used for integrals over contours that are not circles?
5. What is the geometric interpretation of the residue theorem?

Tip: When using the residue theorem, always ensure that you correctly identify the poles within the contour to avoid mistakes in calculating the integral.