We need to evaluate the following contour integral:
\[ \int_C \frac{e^{2z}}{z-4} \, dz \]
where \( C \) is a circle defined by \( |z| = 2 \), oriented counterclockwise. Let's analyze it.
The integrand is structured as \( \frac{f(z)}{z - a} \), with \( f(z) = e^{2z} \), which is analytic (holomorphic) everywhere. The denominator \( z - 4 \) presents a singularity at \( z = 4 \).
Cauchy's Integral Formula applies if \( f(z) \) is analytic inside and on a closed contour \( C \), and \( a \) is inside \( C \):
\[ \int_C \frac{f(z)}{z - a} \, dz = 2\pi i \, f(a) \]
However, in this problem, the singularity \( z = 4 \) is located outside the contour \( C \) because \( |4| > 2 \) and \( C \) is the circle \( |z| = 2 \). Thus, Cauchy's formula is not directly applicable here.
Since the singularity lies outside the contour, the residue at \( z = 4 \) does not influence the integral's value. Applying the residue theorem (or Cauchy's theorem) leads to:
\[ \int_C \frac{e^{2z}}{z-4} \, dz = 0 \]
The contour integral evaluates to:
\[ \boxed{0} \]
Match List-I with List-II and choose the correct option:
| LIST-I (Function) | LIST-II (Value) |
|---|---|
| (A) \( \int_{\gamma} \frac{1}{z-a} \, dz \), where \( \gamma: |z-a|=r, r > 0 \) | (III) \( 2i\pi \) |
| (B) \( \int_{\gamma} \frac{z+2}{z} \, dz \), where \( \gamma: z = 2e^{it}, 0 \le t \le \pi \) | (IV) \( i\pi \) |
| (C) \( \int_{\gamma} \frac{e^{2z}}{(z-1)(z-2)} \, dz \), where \( \gamma: |z|=3 \) | (II) \( 2i\pi(e^4 - e^2) \) |
| (D) \( \int_{\gamma} \frac{z^2 - z + 1}{2(z-1)} \, dz \), where \( \gamma: |z|=2 \) | (I) \( -4 + 2i\pi \) |
Choose the correct answer from the options given below: