Question

The coefficient of \( \frac{1}{z} \) in the Laurent's series expansion of the function \( f(z) = \frac{1{z^2(1-z)} \) about \( z = 0 \), is \dots\dots.

• A. \( -1 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( 2 \)

Hint:

The coefficient of \( \frac{1}{z-a} \) in a Laurent series expansion is equivalent to the residue of the function at that pole. For a pole of order 2, it can also be computed using the limit derivative identity: \(\text{Residue} = \lim_{z \to 0} \frac{d}{dz}\left[z^2 f(z)\right] = \lim_{z \to 0} \frac{d}{dz}\left[\frac{1}{1-z}\right] = \lim_{z \to 0} \frac{1}{(1-z)^2} = 1\).

Answer 1

The Correct Option is C

Concept: A Laurent series expansion represents a complex function inside an annular region about a singularity point \( z = a \). The expansion consists of a principal part (negative power terms) and an analytic part (positive power terms): \[ f(z) = \sum_{n=-\infty}^{\infty} a_n (z-a)^n \] The coefficient of the \( \frac{1}{z-a} \) term (where \( n = -1 \)) is called the Residue of the function at that singular point. For simple functions, this series can be derived using the standard geometric series expansion valid for \( |z| < 1 \): \[ \frac{1}{1-z} = 1 + z + z^2 + z^3 + \dots = \sum_{n=0}^{\infty} z^n \]

Step 1: Expand using Geometric Series
The given function is: \[ f(z) = \frac{1}{z^2(1-z)} \] We isolate the singular term at the origin and expand the remaining part about \( z = 0 \) using the geometric series identity for the region \( 0 < |z| < 1 \): \[ f(z) = \frac{1}{z^2} \cdot \left( \frac{1}{1-z} \right) \] \[ f(z) = \frac{1}{z^2} \cdot \left( 1 + z + z^2 + z^3 + z^4 + \dots \right) \]

Step 2: Distribute the Term into the Series
Multiply \( \frac{1}{z^2} \) across each term inside the parentheses: \[ f(z) = \left( \frac{1}{z^2} \cdot 1 \right) + \left( \frac{1}{z^2} \cdot z \right) + \left( \frac{1}{z^2} \cdot z^2 \right) + \left( \frac{1}{z^2} \cdot z^3 \right) + \dots \] Simplifying the terms: \[ f(z) = \frac{1}{z^2} + \frac{1}{z} + 1 + z + z^2 + \dots \]

Step 3: Identify the Target Coefficient
We need to find the coefficient of the \( \frac{1}{z} \) term in this expansion. Looking at the simplified Laurent series: \[ f(z) = z^{-2} + \mathbf{1} \cdot z^{-1} + 1 + z + z^2 + \dots \] The coefficient of \( z^{-1} \) (which is \( \frac{1}{z} \)) is exactly \( 1 \). This corresponds to Option (C).