LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The logarithmic derivative has residue equal to local order

Statement

Let $f$ be meromorphic on a neighbourhood of $a \in C$ .

If $a$ is a zero of $f$ of order $m \geq 1$ , then $Res ⁣ (\frac{f^{'}}{f}, a) = m .$
If $a$ is a pole of $f$ of order $m \geq 1$ , then $Res ⁣ (\frac{f^{'}}{f}, a) = - m .$

In either case $f^{'} / f$ has a simple pole at $a$ .

Facts & Assumptions

Given: A meromorphic function $f$ on a neighbourhood of $a$ .

[L1]

A holomorphic function has a zero of order $m$ at $a$ exactly when it factors locally as $(z - a)^{m} h (z)$ with $h$ holomorphic and $h (a) \neq 0$ (The order of a zero is the exponent in its local holomorphic factorization).

[L2]

A pole of order $m$ is exactly a point where $1 / f$ extends holomorphically across $a$ and has a zero of order $m$ there (Characterizations of poles).

[L3]

Holomorphic quotients and products obey the usual derivative rules, and a holomorphic function is continuous (Linearity, product, reciprocal, and quotient rules for complex derivatives, Complex differentiability at a point implies continuity there).

Proof

technique · direct

1.1

Suppose first that $a$ is a zero of $f$ of order $m$ . By [L1], on a disc about $a$ one has $f (z) = (z - a)^{m} h (z)$ with $h$ holomorphic and $h (a) \neq 0$ .

givenL1

1.2

Suppose instead that $a$ is a pole of $f$ of order $m$ . By [L2], $1 / f = (z - a)^{m} h$ locally for some holomorphic $h$ with $h (a) \neq 0$ . Shrinking as before, $h$ is nowhere zero.

givenL2L3

2.1

By continuity in [L3], shrink the disc so that $h$ is nowhere zero there. Differentiating the factorization from step 1.1 and dividing by $(z - a)^{m} h (z)$ gives $\frac{f^{'} (z)}{f (z)} = \frac{m}{z - a} + \frac{h^{'} (z)}{h (z)} .$ The second term is holomorphic by [L3], so the residue is $m$ and the pole is simple.

step 1.1L3algebra

3.1

From step 1.2 one has $f (z) = (z - a)^{- m} h (z)^{- 1}$ . Differentiating and dividing by $f$ yields $\frac{f^{'} (z)}{f (z)} = - \frac{m}{z - a} - \frac{h^{'} (z)}{h (z)} .$ Again the second term is holomorphic by [L3], so the residue is $- m$ and the pole is simple.

step 1.2L3algebra∎

Depends on

Used by

The argument principle for an admissible null-homologous cycle Theorem
The weighted argument principle Theorem

Dependency tree · two levels

23 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

R. W. Howell and J. H. Mathews, Complex Analysis, §8.7 (standard reference, not scraped)
J. Lebl, Guide to Cultivating Complex Analysis, §5.4 (standard reference, not scraped)