Residue of Quotient
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Theorem
Let $f$ and $g$ be functions holomorphic on some region containing $a$.
Let $g$ have a zero of multiplicity $1$ at $a$.
Then:
- $\Res {\dfrac f g} a = \dfrac {\map f a} {\map {g'} a}$
This article, or a section of it, needs explaining. In particular: definition of "Res", by a link to Residue You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Proof
As $g$ has a zero of multiplicity $1$ at $a$, $\dfrac f g$ has a simple pole at $a$ by definition.
So:
\(\ds \Res {\dfrac f g} a\) | \(=\) | \(\ds \lim_{z \mathop \to a} \paren {z - a} \frac {\map f z} {\map g z}\) | Residue at Simple Pole | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{z \mathop \to a} \frac {\map f z} {\frac {\map g z - \map g a} {z - a} }\) | $\map g a = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\lim_{z \mathop \to a} \map f z} {\lim_{z \mathop \to a} \frac {\map g z - \map g a} {z - a} }\) | Quotient Rule for Limits of Complex Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map f a} {\map {g'} a}\) | Definition of Derivative of Complex Function |
$\blacksquare$