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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

1.1givenL1

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 $h (a) \neq 0$ .

1.2givenL2L3

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.

2.1step 1.1L3algebra

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.

3.1step 1.2L3algebra∎

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.

The logarithmic derivative has residue equal to local order

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources