Finding the $\operatorname{Res}(\frac gh, a)$ with $h$ and $g$ analytic at $a$ I'm having trouble with a problem form my complex analysis class about the residue of the quotient of two analytic functions. It states that if you have $g$ and $h$, both analytic functions at a point $a$, such that $g(a)\neq0$ and $h$ has $a$ as a zero of order 2, then: 
$\operatorname{Res}(\frac gh, a) = 2\frac{g'(a)}{h''} - \frac{2}{3}\frac{g(a)h'''(a)}{(h''(a))^2}$
I'm guessing this follows from Taylor, where $f=\frac{h}{g}$ can be expressed as:
$f(z)= \sum_{n=0}^{\infty} \frac{f^n(z_0)}{n!} (z-z_0)^n$
maybe also using the fact that since $h$ is analytic at and has $a$ as a zero then it can be expressed like $h=(z-a)^2 h_1(z)$, where $h_1(z)$ is analytic at $a$ and $h_1(a)\neq 0$.
I'm not seeing how to "connect the dots" here or even if these are "dots" that can be connected in this case, so any hint towards a correct idea would be appreciated.

You have$$g(z)=a_0+a_1(z-a)+\cdots$$and$$h(z)=b_2(z-a)^2+b_3(z-a)^3+\cdots,$$with $a_0,b_2\neq0$. Then the Laurent series of $\frac gh$ around $0$ is of the type$$\frac{c_{-2}}{(z-a)^2}+\frac{c_{-1}}{z-a}+c_0+\cdots$$and you want to determine $c_{-1}$. Since$$\frac{a_0+a_1(z-a)+\cdots}{b_2(z-a)^2+b_3(z-a)^3+\cdots}=\frac{c_{-2}}{(z-a)^2}+\frac{c_{-1}}{z-a}+c_0+\cdots,$$you know that\begin{align}a_0+a_1(z-a)+\cdots&=\left(b_2(z-a)^2+b_3(z-a)^3+\cdots\right)\left(\frac{c_{-2}}{(z-a)^2}+\frac{c_{-1}}{z-a}+c_0+\cdots\right)\\&=(b_2+b_3(z-a)+\cdots)(c_{-2}+c_{-1}(z-1)+\cdots)\\&=b_2c_{-2}+(b_2c_{-1}+b_3c_{-2})(z-a)+\cdots\end{align}Therefore, you have$$\left\{\begin{array}{l}b_2c_{-2}=a_0\\b_2c_{-1}+b_3c_{-2}=a_1.\end{array}\right.$$From the first equation, you get that $c_{-2}=\dfrac{a_0}{b_2}$. Using this, from the second equation you get that$$c_{-1}=\frac{a_1}{b_2}-\frac{b_3a_0}{{b_2}^2}.$$All that is left is to use these facts:


*

*$\displaystyle a_0=g(a)$;

*$\displaystyle a_1=g'(a)$;

*$\displaystyle b_2=\frac{h''(a)}2$;

*$\displaystyle b_3=\frac{h'''(a)}6$.

