Residue of complex function involving logarithm

Hi,

I need some help calculating the residues of the following function at z = +-i:

(\frac{lnz}{1+z^2})^2

At first glance I thought the function did not contain any simple poles as the denominator looks second order in (z+i) and (z-i). However, after going on wolfram, the Laurent series at z=i shows a non-zero coefficient for the 1/(z-i) term (http://www.wolframalpha.com/input/?i=laurent+series+%28ln%28z%29%2F%281%2Bz%5E2%29%29%5E2+at+z%3Di).

The residues are also used in this wiki page on example V: http://en.wikipedia.org/wiki/Methods_of_contour_integration.

How do I calculate these residues? I tried the standard proceedures but get nowhere...

I've tried:

lim_{z\to i} (z-i)f(z)

and

\frac{q(z_o)}{p'(z_o)}

and still can't get it...

Reply 1

Farhan.Hanif93

Original post by soutioirsim

However, after going on wolfram, the Laurent series at z=i shows a non-zero coefficient for the 1/(z-i) term.

The poles are still of order 2. The non-vanishing of the

\dfrac{1}{z-i}

term is of no consequence given that the

\dfrac{1}{(z-i)^2}

term has a non-zero coefficient.

z=a

is a pole of order

N

f(z)

with Laurent expansion

\displaystyle\sum_{n=-\infty}^{\infty}c_n(z-a)^n \iff c_N \not= 0 \ \text{and} \ c_n = 0 \forall n< N

How do I calculate these residues? I tried the standard proceedures but get nowhere...

Take a suitable limit, now that you know the poles are of order 2.

Reply 2

soutioirsim

Original post by Farhan.Hanif93

The poles are still of order 2. The non-vanishing of the

\dfrac{1}{z-i}

term is of no consequence given that the

\dfrac{1}{(z-i)^2}

term has a non-zero coefficient.

z=a

is a pole of order

N

f(z)

with Laurent expansion

\displaystyle\sum_{n=-\infty}^{\infty}c_n(z-a)^n \iff c_N \not= 0 \ \text{and} \ c_n = 0 \forall n< N

Take a suitable limit, now that you know the poles are of order 2.

Thank you! One question however, I thought that Cauchy's residue theorem only works with simple poles? How is it in this case we are calculating residues for 2 order poles?

Reply 3

davros

Original post by soutioirsim

Thank you! One question however, I thought that Cauchy's residue theorem only works with simple poles? How is it in this case we are calculating residues for 2 order poles?

Why do you think that?

There is a fundamental result called Cauchy's integral formula which tells you the value of f(a) in terms of the integral of f(z)/(z - a), but that formula can be differentiated to give a more general result, and you should have a formula for calculating residues for an nth order pole (the formula you quoted in your first post was only true for calculating the residue at a simple pole!)

Reply 4

Farhan.Hanif93

Original post by soutioirsim

Thank you! One question however, I thought that Cauchy's residue theorem only works with simple poles? How is it in this case we are calculating residues for 2 order poles?

Cauchy's residue theorem applies just as well to all isolated singularities with a calculable residue - the residues just get more difficult to compute, in general.

See here.

Reply 5

soutioirsim

Original post by Farhan.Hanif93

Cauchy's residue theorem applies just as well to all isolated singularities with a calculable residue - the residues just get more difficult to compute, in general.

See here.

Original post by davros