Answer: Two Laurent Series for One Function

Key Idea / Intuition

A Laurent series is not just a property of a function — it is a property of a function on an annulus. The same meromorphic function can have completely different Laurent series in different annular regions, because each region "sees" different poles as being "inside" vs "outside." The coefficients are determined by residues and integrals that depend essentially on which singularities are enclosed. This is the heart of why Laurent series are attached to domains, not just to functions.

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Formal Proof / Solution

Setup: Partial Fractions

First, decompose $f$ via partial fractions: $f(z) = \frac{1}{z(z-1)} = \frac{-1}{z} + \frac{1}{z-1}.$

(Check: $\frac{-1}{z} + \frac{1}{z-1} = \frac{-(z-1) + z}{z(z-1)} = \frac{1}{z(z-1)}$. ✓)

The singularities are at $z = 0$ (pole) and $z = 1$ (pole).

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Part (a): Laurent Series on $0 < |z| < 1$

In this region, $|z| < 1$, so $\frac{1}{z-1} = \frac{-1}{1-z}$ can be expanded as a geometric series: $\frac{1}{z-1} = -\frac{1}{1-z} = -\sum_{n=0}^{\infty} z^n, \quad |z| < 1.$

Therefore: $f(z) = -\frac{1}{z} + \left(-\sum_{n=0}^{\infty} z^n\right) = -\frac{1}{z} - 1 - z - z^2 - z^3 - \cdots$

$\boxed{f(z) = -\frac{1}{z} - \sum_{n=0}^{\infty} z^n, \quad 0 < |z| < 1.}$

This has a simple pole at $z=0$ (the $-1/z$ term), as expected.

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Part (b): Laurent Series on $|z| > 1$

In this region, $|z| > 1$, so $\frac{1}{z-1}$ should be expanded in powers of $1/z$: $\frac{1}{z-1} = \frac{1}{z}\cdot\frac{1}{1 - 1/z} = \frac{1}{z}\sum_{n=0}^{\infty} \frac{1}{z^n} = \sum_{n=0}^{\infty} \frac{1}{z^{n+1}}, \quad |z| > 1.$

Also, $\frac{-1}{z}$ is already a power of $1/z$. Combining: $f(z) = -\frac{1}{z} + \sum_{n=0}^{\infty} \frac{1}{z^{n+1}} = -\frac{1}{z} + \frac{1}{z} + \frac{1}{z^2} + \frac{1}{z^3} + \cdots$

The $-1/z$ and $+1/z$ cancel!

$\boxed{f(z) = \sum_{n=2}^{\infty} \frac{1}{z^n} = \frac{1}{z^2} + \frac{1}{z^3} + \frac{1}{z^4} + \cdots, \quad |z| > 1.}$

This series has no negative powers beyond $z^{-2}$ — in particular, the residue at $\infty$ is zero, which is consistent with $f(z) \to 0$ as $|z| \to \infty$.

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Part (c): Conceptual Reason They Differ

The Laurent series on an annulus $r < |z| < R$ is unique — there is exactly one such series converging there. The two annuli $0 < |z| < 1$ and $|z| > 1$ are separated by the singularity at $z = 1$.

• On $0 < |z| < 1$: the singularity at $z=1$ is outside the disk, so $\frac{1}{z-1}$ expands in non-negative powers of $z$.
• On $|z| > 1$: the singularity at $z=1$ is inside the circle, so $\frac{1}{z-1}$ must be expanded in negative powers of $z$.

In the language of the Cauchy integral formula: the Laurent coefficients $c_n = \frac{1}{2\pi i}\oint_{|z|=r} f(z)z^{-n-1}\,dz$ depend on $r$. Crossing the singularity at $|z|=1$ changes which poles are enclosed, changing all the coefficients.

The two series are genuinely different functions of their respective variables — they just happen to represent the same meromorphic function in their respective domains of validity.