Complex Analysis Residue question

morganfor · #1 $Old$ June 4th, 2009, 01:22 PM

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!

shawsend · #2 $Old$ June 4th, 2009, 05:13 PM

Quote:

Originally Posted by morganfor $View Post$

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.

Random Variable · #3 $Old$ June 4th, 2009, 05:21 PM

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.

morganfor · #4 $Old$ June 5th, 2009, 11:22 AM

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

shawsend · #5 $Old$ June 5th, 2009, 03:26 PM

Quote:

Originally Posted by shawsend $View Post$

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 $b_1$
$b_1=-2/3 \frac{f'''(z_0)}{(f''(z_0))^2}$

Random Variable · #6 $Old$ June 5th, 2009, 03:38 PM

Quote:

Originally Posted by shawsend $View Post$

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 $f'(z_{0}) = 0$ ?

shawsend · #7 $Old$ June 5th, 2009, 04:02 PM

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 $f'(z_0)=0$ .

Random Variable · #8 $Old$ June 5th, 2009, 07:35 PM

Quote:

Originally Posted by shawsend $View Post$

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.