Question

If C is the positively oriented circle represented by \( |z|=2 \), then \( \int_C \frac{e^{2z}}{z-4} dz \) is:

• A. \( \frac{2\pi i}{3} \)
• B. \( \pi i \)
• C. \( \frac{4\pi i}{3} \)
• D. 0

Hint:

When evaluating a contour integral, the very first step is to locate the singularities (poles) of the integrand and check if they are inside, outside, or on the contour. If all singularities are outside, the integral is zero by Cauchy's Integral Theorem.

Answer 1

The Correct Option is D

We are tasked with evaluating the contour integral:

\[ \int_C \frac{e^{2z}}{z-4} \, dz \]

where \( C \) is the positively oriented circle given by \( |z| = 2 \). Let's proceed step by step.

Step 1: Recognize the Form of the Integral

The integrand is of the form \( \frac{f(z)}{z - a} \), where \( f(z) = e^{2z} \) is analytic (holomorphic) everywhere. The denominator \( z - 4 \) has a singularity at \( z = 4 \).

Step 2: Apply Cauchy's Integral Formula

Cauchy's Integral Formula states that if \( f(z) \) is analytic inside and on a closed contour \( C \), and \( a \) is a point inside \( C \), then:

\[ \int_C \frac{f(z)}{z - a} \, dz = 2\pi i \, f(a) \]

However, in our case, the singularity \( z = 4 \) lies outside the contour \( C \) (since \( |4| > 2 \), and \( C \) is the circle \( |z| = 2 \)). Therefore, we cannot directly apply Cauchy's formula here.

Step 3: Conclusion Based on Residues

Since the singularity lies outside the contour, the residue of the integrand at \( z = 4 \) does not contribute to the integral. Therefore, by the residue theorem (or Cauchy's theorem), we conclude that:

\[ \int_C \frac{e^{2z}}{z-4} \, dz = 0 \]

Final Answer:

The value of the contour integral is:

\[ \boxed{0} \]