Closed-form expression for residue of a pole of order 2

Question

Suppose $f(z)/g(z)$ has a pole of order 1 at $c$. Then its residue at $c$ is $f(c)/g'(c)$. I want a formula like this of a quotient for a pole of order 2. I know it's $\lim_{z\to c} \frac{d}{dz}[(z-c)^2\frac{f(z)}{g(z)}]$, but I want a more concise expression in terms of the derivatives of $g$ and/or $f$.

Computing $\frac{d}{dz}[(z-c)^2\frac{f(z)}{g(z)}]$ and noting that $g(c) = 0$ by assumption, we get

$$ 2(z - c)\frac{f}{g} + (z-c)^2\frac{f'g - fg'}{g^2}\\ = 2(z - c)\frac{f}{g} + \frac{f'g - fg'}{\left(\frac{g(z) - g(c)}{z-c}\right)^2} \\ \to 2\frac{f(c)}{g'(c)} + \frac{f'(c)\cdot0 - f(c)g'(c)}{g'(c)^2} \\ = \frac{f(c)}{g'(c)} $$

But this is the same expression for the first-order pole, and I know this is wrong. Where did I make a mistake? And what's the right expression?

Answer 1

I presume $f(c)\ne0$ but $g$ has a double zero at $z$. Then $$f(z)=a_0+a_1w+a_2w^2+\cdots$$ and $$g(z)=b_2w^2+b_3w^3+\cdots$$ where $w=z-c$. Then $$\frac{f(z)}{g(z)}=\frac{1}{w^2}\frac{a_0+a_1w+\cdots}{b_2+b_3w+\cdots} =\frac1{b_2w^2}(a_0+a_1w+\cdots)(1-(b_3/b_2)w+\cdots) =\frac{a_0+(a_1-a_0b_3/b_2)w+\cdots}{b_2w^2}$$ and so the residue is $$\frac{a_1-a_0b_3/b_2}{b_2} =\frac{a_1b_2-a_0b_3}{b_2^2}=\frac{f'(c)g''(c)/2-f(c)g'''(c)/6}{g''(c)^2/4} =\frac{6f'(c)g''(c)-2f(c)g'''(c)}{3g''(c)^2}.$$

Answer 2

It is NOT true that $$ 2(z - c)\frac{f}{g} + \frac{f'g - fg'}{\left(\frac{g(z) - g(c)}{z-c}\right)^2} \longrightarrow 2\frac{f(c)}{g'(c)} + \frac{f'(c)\cdot0 - f(c)g'(c)}{g'(c)^2} $$ because $$ \frac{f(c)}{g'(c)} $$ does not exist, since $z=c$ is a double root for $g$.