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:

$\Res {\dfrac f g} a = \dfrac {\map f a} {\map {g'} a}$




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

$\blacksquare$