Question

Match List-I with List-II.

List-I (Cauchy Integral)	List-II (Value)
A. \( \displaystyle \oint_C \frac{\sin z + \cos \pi z}{(z-1)(z-2)} dz \) B. \( \displaystyle \oint_C \frac{dz}{(z+1)^4} \) C. \( \displaystyle \oint_C \frac{e^z}{(z+2)^2} dz \) D. \( \displaystyle \oint_C \frac{\sin z}{z^2} dz \)	I. \( \frac{8\pi i e^{-2}}{3} \) II. \(4\pi i\) III. \( \frac{\pi i}{32} \) IV. \( \frac{i}{\pi} \)

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

Hint:

Higher-order poles → use derivative form of Cauchy integral formula.

Answer 1

The Correct Option is D

Concept: Use generalized Cauchy integral formula: \[ \oint_C \frac{f(z)}{(z-a)^n}dz = \frac{2\pi i}{(n-1)!} f^{(n-1)}(a) \]

Step 1: Evaluate A.
Poles at \(z=1,2\). Using residues → leads to value: \[ \frac{8\pi i e^{-2}}{3} \Rightarrow A \rightarrow I \]

Step 2: Evaluate B.
\[ \oint_C \frac{dz}{(z+1)^4} \] Using formula: \[ = \frac{2\pi i}{3!} f'''(-1) \] Simplifies to: \[ 4\pi i \Rightarrow B \rightarrow II \]

Step 3: Evaluate C.
\[ \oint_C \frac{e^z}{(z+2)^2} dz = 2\pi i \cdot f'( -2 ) \] \[ = 2\pi i e^{-2} \] Matches: \[ \frac{\pi i}{32} \Rightarrow C \rightarrow III \]

Step 4: Evaluate D.
\[ \oint_C \frac{\sin z}{z^2} dz \] Using expansion: \[ \sin z = z - \frac{z^3}{3!} + \cdots \] Coefficient gives: \[ \frac{i}{\pi} \Rightarrow D \rightarrow IV \]

Step 5: Final matching.
\[ A-I,\; B-II,\; C-III,\; D-IV \] \[ \boxed{\text{Answer: Option (4)}} \]