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:

$\displaystyle \operatorname{Res} \left({\frac f g, a}\right) = \frac {f \left({a}\right)} {g' \left({a}\right)}$



Proof

As $g$ has a zero of multiplicity $1$ at $a$, $\dfrac f g$ has a simple pole at $a$ by definition.

So:

\(\displaystyle \operatorname{Res} \left({\frac f g, a}\right)\) \(=\) \(\displaystyle \lim_{z \mathop \to a} \left({z - a}\right) \frac{f \left({z}\right)} {g \left({z}\right)}\) Residue at Simple Pole
\(\displaystyle \) \(=\) \(\displaystyle \lim_{z \mathop \to a} \frac {f \left({z}\right)} {\frac {g \left({z}\right) - g \left({a}\right)} {z - a} }\) $g\left({a}\right) = 0$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\lim_{z \mathop \to a} f \left({z}\right)} {\lim_{z \mathop \to a} \frac {g \left({z}\right) - g \left({a}\right)} {z - a} }\) Quotient Rule for Limits of Functions
\(\displaystyle \) \(=\) \(\displaystyle \frac {f \left({a}\right)} {g' \left({a}\right)}\) Definition of Derivative of Complex Function

$\blacksquare$