Question

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:

• A. A - I, B - II, C - III, D - IV
• B. A - I, B - III, C - II, D - IV
• C. A - III, B - I, C - II, D - IV
• D. A - III, B - IV, C - I, D - II

Hint:

When evaluating complex integrals, first check if the contour is closed. If it is, check the singularities inside. For a single pole, Cauchy's Integral Formula is efficient. For multiple poles, the Residue Theorem is necessary. If the contour is not closed, direct parameterization is the standard approach.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question requires the evaluation of four different complex contour integrals using various methods of complex analysis, including Cauchy's Integral Formula, parameterization, and the Residue Theorem.

Step 2: Detailed Explanation:
A. The integral is \( \int_{\gamma} \frac{1}{z-a} dz \) over a simple closed contour \(|z-a|=r\) enclosing the point \(a\). Let \(f(z) = 1\). By Cauchy's Integral Formula, \( \int_{\gamma} \frac{f(z)}{z-a} dz = 2\pi i f(a) \). \[ \int_{\gamma} \frac{1}{z-a} dz = 2\pi i \cdot (1) = 2i\pi \] Therefore, A matches III.
B. The contour is a semi-circle from \(z=2\) to \(z=-2\), so it's not a closed path. We must evaluate by parameterization. Let \( z = 2e^{it} \), so \( dz = 2ie^{it} dt \). The limits are \( t \in [0, \pi] \). \[ \int_{\gamma} \frac{z+2}{z} dz = \int_{\gamma} \left(1 + \frac{2}{z}\right) dz = \int_{0}^{\pi} \left(1 + \frac{2}{2e^{it}}\right) (2ie^{it}) dt = \int_{0}^{\pi} (1 + e^{-it}) (2ie^{it}) dt \] \[ = \int_{0}^{\pi} (2ie^{it} + 2i) dt = \left[ \frac{2ie^{it}}{i} + 2it \right]_{0}^{\pi} = \left[ 2e^{it} + 2it \right]_{0}^{\pi} \] \[ = (2e^{i\pi} + 2i\pi) - (2e^{i0} + 0) = (2(-1) + 2i\pi) - 2 = -4 + 2i\pi \] Therefore, B matches I.
C. The contour is \(|z|=3\), which is a circle that encloses both simple poles at \(z=1\) and \(z=2\). We use the Residue Theorem. Let \( f(z) = \frac{e^{2z}}{(z-1)(z-2)} \). \[ \text{Res}(f, 1) = \lim_{z \to 1} (z-1)f(z) = \lim_{z \to 1} \frac{e^{2z}}{z-2} = \frac{e^2}{1-2} = -e^2 \] \[ \text{Res}(f, 2) = \lim_{z \to 2} (z-2)f(z) = \lim_{z \to 2} \frac{e^{2z}}{z-1} = \frac{e^4}{2-1} = e^4 \] By the Residue Theorem, the integral is \( 2\pi i (\text{Sum of residues}) = 2\pi i (-e^2 + e^4) = 2i\pi(e^4 - e^2) \). Therefore, C matches II.
D. The contour is \(|z|=2\), which encloses the simple pole at \(z=1\). We use Cauchy's Integral Formula. \[ \int_{\gamma} \frac{z^2-z+1}{2(z-1)} dz = \frac{1}{2} \int_{\gamma} \frac{z^2-z+1}{z-1} dz \] Let \(g(z) = z^2 - z + 1\). The integral is \( \frac{1}{2} [2\pi i \cdot g(1)] \). \[ = \pi i \cdot (1^2 - 1 + 1) = i\pi \] Therefore, D matches IV.

Step 3: Final Answer:
The correct matching is A-III, B-I, C-II, D-IV. This corresponds to option (C).