if f has a zero of order 2 at z0 and A=f''(z0) and B=f'''(z0) and h(z)=1/f(z), how does one find a formula for Res(h;z0) in terms of A and B?
thanks!
Just expand them in terms of Laurent series:
$\displaystyle \frac{g(z)}{h(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{(z-z_0)}+\cdots$
$\displaystyle h(z)=\frac{h''(z_0)}{2}(z-z_0)^2+\frac{h'''(z_0)}{6}(z-z_0)^2$
Therefore $\displaystyle g(z)=h(z)\left[\frac{b_2}{(z-z_0)^2}+\cdots \right]$
Multiply it out, take derivatives, equate coefficients and solve for $\displaystyle b_1$
I'd recommend you find "Basic Complex Analysis" by Marsden and Hoffman. It's a nice read on basic stuff and it goes over this derivation.
You could try the following:
$\displaystyle Res[h,z_{0}] = \lim_{z \to z_{0}} \frac {d}{dz} (z-z_{0})^{2} h(z) $
$\displaystyle = \lim_{z \to z_{0}} \frac {d}{dz} (z-z_{0})^{2} \frac {1}{f(z)} $
then using the product rule
$\displaystyle = \lim_{z \to z_{0}} \Big(2(z-z_{0}) \frac {1}{f(z)} + (z-z_{0})^{2} \frac {-1}{[f(z)]^2} f'(z) \Big) $
And then since the forms of both terms are indeterminate, apply L'Hospital's rule. But multiple applications will be required.
I'm sorry my $\displaystyle h(z)$ conflicts with your $\displaystyle h(z)$. Mine was just general terms. You have:
$\displaystyle \frac{1}{f(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{z-z_0}+a_0+\cdots$
and:
$\displaystyle f(z)=\frac{f''(z_0)}{2}(z-z_0)^2+\frac{f'''(z_0)}{6}(z-z0)^3+\cdots$
and therefore
$\displaystyle 1=\left[\frac{f''(z_0)}{2}+\frac{f'''(z_0)}{6}(z-z_0)+\cdots\right]\cdot\left[b_2+b_1(z-z_0)+a_1(z-z_0)^2+\cdots\right]$
so that
$\displaystyle 1=\frac{f''(z_0}{2} b_2+\left(\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}\right)(z-z_0)+\cdots$
Equating coefficients:
$\displaystyle 1=\frac{b_2 f''(z_0)}{2}$
and:
$\displaystyle 0=\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}$
Solving for $\displaystyle b_1$
$\displaystyle b_1=-2/3 \frac{f'''(z_0)}{(f''(z_0))^2}$
If $\displaystyle f(z)$ has a zero of order 2 at $\displaystyle z_0$, then $\displaystyle f(z)$ can be written as:
$\displaystyle f(z)=a_2(z-z_0)^2+a_3(z-z_0)^3+\cdots$
but that's the Taylor series which is unique so that:
$\displaystyle a_2=\frac{f''(z_0)}{2}$
$\displaystyle a_3=\frac{f'''(z_0)}{6}$
and so on. That means the first two terms of the Taylor series must be zero or $\displaystyle f(z_0)=0$ and $\displaystyle f'(z_0)=0$.