QB_PH_004: Linear Algebra & Group Theory

Consider the complex function (z) = rac{1}{z(z - 1)(z - 2)}$. What is the residue of (z)$ at the pole = 10

Correct Answer

### 1. Conceptual Framework & Core Principles For a meromorphic function (z)$ with a simple pole at = z_0$, the residue is given by: 12696 ext{Res}(f, z_0) = \lim_{z o z_0} (z - z_0) f(z)12696 ### 2. Analytical & Mathematical Derivation Here, (z) = rac{1}{z(z - 1)(z - 2)}$. The singularities are simple poles at = 0$, = 1$, and = 2$. Evaluating at = 1$: 12696 ext{Res}(f, 1) = \lim_{z o 1} (z - 1) rac{1}{z(z - 1)(z - 2)} = \lim_{z o 1} rac{1}{z(z - 2)}12696 Substituting = 1$: 12696 ext{Res}(f, 1) = rac{1}{1(1 - 2)} = rac{1}{-1} = -112696 ### 3. Formula Breakdown - Simple pole residue: $ ext{Res}(f, z_0) = \lim_{z o z_0} (z - z_0) f(z)$ - Value: $ rac{1}{1 \cdot (-1)} = -1$. ### 4. Student Traps & Exam Tips **Common Trap:** Forgetting the sign in = -1$, leading students to erroneously choose 0$.