GATE PH 2025 • Q21
MCQ • 1 Mark
Section 1: Mathematical Physics
Complex analysis
Subtopic: Residue Theorem & Contour Integration
Question Statement:
Consider the function f(z) =
1
z2(z-2)3 of a complex variable z. The residues of
the function at z = 0 and z = 2, respectively, are
Options:
-
(A) -
3
8 and
3
8
-
(B) 3
8 and -
3
16
-
(C) -
3
16 and
3
16
-
(D) -
3
8 and
3
16
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.21)**
**1. Complex Analysis & Cauchy Residue Theorem**:
- Given rational complex function:
$f(z) = \frac{1}{z^2(z - 2)^3}$
- The function has:
- A pole of order $m = 2$ at $z = 0$.
- A pole of order $m = 3$ at $z = 2$.
**2. Step-by-Step Residue Evaluation**:
- **Residue at $z = 0$ (Pole of order 2)**:
$\text{Res}(f, 0) = \lim_{z \to 0} \frac{1}{(2-1)!} \frac{d}{dz}\left[z^2 f(z)\right] = \lim_{z \to 0} \frac{d}{dz}\left[(z - 2)^{-3}\right]$
$\text{Res}(f, 0) = \lim_{z \to 0} \left[-3(z - 2)^{-4}\right] = -\frac{3}{(-2)^4} = -\frac{3}{16}$
- **Residue at $z = 2$ (Pole of order 3)**:
- Method 1 (Direct Derivative):
$\text{Res}(f, 2) = \frac{1}{2!} \lim_{z \to 2} \frac{d^2}{dz^2}\left[(z - 2)^3 f(z)\right] = \frac{1}{2} \lim_{z \to 2} \frac{d^2}{dz^2}[z^{-2}]$
$\frac{d}{dz}[z^{-2}] = -2z^{-3}, \quad \frac{d^2}{dz^2}[z^{-2}] = 6z^{-4}$
$\text{Res}(f, 2) = \frac{1}{2} \left(\frac{6}{2^4}\right) = \frac{3}{16}$
- Method 2 (Residue at Infinity):
Because $f(z) \sim \frac{1}{z^5}$ as $z \to \infty$, the sum of all residues in the extended complex plane vanishes:
$\text{Res}(f, 0) + \text{Res}(f, 2) = 0 \implies \text{Res}(f, 2) = -\text{Res}(f, 0) = -\left(-\frac{3}{16}\right) = +\frac{3}{16}$
**3. Official Verification**:
The verified official IIT answer is **Option (C)** (`-3/16 and 3/16`).
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