LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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At a simple pole the residue is the limit of (z-a)f(z)

Statement

If $a$ is a simple pole of $f$ , then

$Res (f, a) = \lim_{z \to a} (z - a) f (z) .$

Facts & Assumptions

Given: A simple pole of $f$ at $a$ .

[L1]

A simple pole is a pole of order $1$ (Simple poles).

[L2]

If $a$ is a pole of order $1$ , then $(z - a) f (z)$ extends holomorphically across $a$ with a nonzero value there (Characterizations of poles).

[L3]

The residue is the coefficient of $(z - a)^{- 1}$ in the Laurent expansion (The residue of an isolated singularity).

[L4]

Holomorphic functions are continuous (Complex differentiability at a point implies continuity there).

Proof

technique · direct

1.1

By [L1] and [L2], $g (z) : = (z - a) f (z)$ extends holomorphically across $a$ ; write the extension again as $g$ , so $g (a)$ is defined and $g (z) \to g (a)$ by [L4].

L1L2L4

1.2

The Laurent expansion of $f$ is $f (z) = c_{- 1} (z - a)^{- 1} + \sum_{n \geq 0} c_{n} (z - a)^{n}$ , because a simple pole has no terms $(z - a)^{- m}$ with $m \geq 2$ . Multiplying by $(z - a)$ gives $g (z) = c_{- 1} + \sum_{n \geq 0} c_{n} (z - a)^{n + 1}$ , so $g (a) = c_{- 1} = Res (f, a)$ by [L3].

L2L3algebra

2.1

Combining steps 1.1 and 1.2 gives $\lim_{z \to a} (z - a) f (z) = g (a) = Res (f, a)$ .

step 1.1step 1.2∎

Depends on

Used by

Residues of p over q at a simple zero of q Corollary

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jean-Baptiste Campesato, MAT334 course page and notes index (standard reference, not scraped)