# Thread: Complex Analysis Residue question

1. ## Complex Analysis Residue question

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!

2. Originally Posted by morganfor
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:

$\frac{g(z)}{h(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{(z-z_0)}+\cdots$

$h(z)=\frac{h''(z_0)}{2}(z-z_0)^2+\frac{h'''(z_0)}{6}(z-z_0)^2$

Therefore $g(z)=h(z)\left[\frac{b_2}{(z-z_0)^2}+\cdots \right]$

Multiply it out, take derivatives, equate coefficients and solve for $b_1$

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.

3. You could try the following:

$Res[h,z_{0}] = \lim_{z \to z_{0}} \frac {d}{dz} (z-z_{0})^{2} h(z)$

$= \lim_{z \to z_{0}} \frac {d}{dz} (z-z_{0})^{2} \frac {1}{f(z)}$

then using the product rule

$= \lim_{z \to z_{0}} \Big(2(z-z_{0}) \frac {1}{f(z)} + (z-z_{0})^{2} \frac {-1}{[f(z)]^2} f'(z) \Big)$

And then since the forms of both terms are indeterminate, apply L'Hospital's rule. But multiple applications will be required.

4. Thank you!
Shawsend, I'm just confused about where the g(z) came from and what it is - do you mean f(z)?

5. Originally Posted by shawsend
Just expand them in terms of Laurent series:

$\frac{g(z)}{h(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{(z-z_0)}+\cdots$

$h(z)=\frac{h''(z_0)}{2}(z-z_0)^2+\frac{h'''(z_0)}{6}(z-z_0)^2$

Therefore $g(z)=h(z)\left[\frac{b_2}{(z-z_0)^2}+\cdots \right]$

Multiply it out, take derivatives, equate coefficients and solve for $b_1$

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.
I'm sorry my $h(z)$ conflicts with your $h(z)$. Mine was just general terms. You have:
$\frac{1}{f(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{z-z_0}+a_0+\cdots$
and:
$f(z)=\frac{f''(z_0)}{2}(z-z_0)^2+\frac{f'''(z_0)}{6}(z-z0)^3+\cdots$
and therefore
$1=\left[\frac{f''(z_0)}{2}+\frac{f'''(z_0)}{6}(z-z_0)+\cdots\right]\cdot\left[b_2+b_1(z-z_0)+a_1(z-z_0)^2+\cdots\right]$
so that
$1=\frac{f''(z_0}{2} b_2+\left(\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}\right)(z-z_0)+\cdots$
Equating coefficients:
$1=\frac{b_2 f''(z_0)}{2}$
and:
$0=\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}$
Solving for $b_1$
$b_1=-2/3 \frac{f'''(z_0)}{(f''(z_0))^2}$

6. Originally Posted by shawsend
I'm sorry my $h(z)$ conflicts with your $h(z)$. Mine was just general terms. You have:
$\frac{1}{f(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{z-z_0}+a_0+\cdots$
and:
$f(z)=\frac{f''(z_0)}{2}(z-z_0)^2+\frac{f'''(z_0)}{6}(z-z0)^3+\cdots$
and therefore
$1=\left[\frac{f''(z_0)}{2}+\frac{f'''(z_0)}{6}(z-z_0)+\cdots\right]\cdot\left[b_2+b_1(z-z_0)+a_1(z-z_0)^2+\cdots\right]$
so that
$1=\frac{f''(z_0}{2} b_2+\left(\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}\right)(z-z_0)+\cdots$
Equating coefficients:
$1=\frac{b_2 f''(z_0)}{2}$
and:
$0=\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}$
Solving for $b_1$
$b_1=-2/3 \frac{f'''(z_0)}{(f''(z_0))^2}$
This might be a stupid question, but how do we know that $f'(z_{0}) = 0$ ?

7. If $f(z)$ has a zero of order 2 at $z_0$, then $f(z)$ can be written as:

$f(z)=a_2(z-z_0)^2+a_3(z-z_0)^3+\cdots$

but that's the Taylor series which is unique so that:

$a_2=\frac{f''(z_0)}{2}$

$a_3=\frac{f'''(z_0)}{6}$

and so on. That means the first two terms of the Taylor series must be zero or $f(z_0)=0$ and $f'(z_0)=0$.

8. Originally Posted by shawsend
If $f(z)$ has a zero of order 2 at $z_0$, then $f(z)$ can be written as:

$f(z)=a_2(z-z_0)^2+a_3(z-z_0)^3+\cdots$

but that's the Taylor series which is unique so that:

$a_2=\frac{f''(z_0)}{2}$

$a_3=\frac{f'''(z_0)}{6}$

and so on. That means the first two terms of the Taylor series must be zero or $f(z_0)=0$ and $f'(z_0)=0$.
But basically what you're using is the fact that if $z_{0}$ is a zero of $f(z)$ of order k, then $f(z{o}), f'(z_{o}), ... , f^{(k-1)}(z_{0}) = 0$, right?

I don't why I didn't realize that before.