# Calculating Laurent series of complex functions

• MatinSAR
MatinSAR
Homework Statement
Calculate laurent series of following function.
Relevant Equations
series expansion formulas.
1. ##f(z)=\dfrac {\sin z}{z- \pi}## at ##z=\pi## : $$\dfrac {\sin z}{z- \pi}=\dfrac {\sin(\pi +z- \pi)}{z- \pi}=\dfrac {- \sin(z- \pi)}{z- \pi}=\dfrac {-1}{z- \pi} \sum_{n=0}^\infty \dfrac {(-1)^n (z- \pi)^{2n}}{(2n+1)!}$$My answer has extra ##\dfrac {-1}{z- \pi} ## according to a calculator. Am I wrong?

2. Find the residue of ##f(z)=\dfrac {\sin z} {z^4}## at ##z=0## :
I wrote laurent series ot this. It doesn't have ##1/z##. I think the residue is ##0##. Is it it true?

3. Calculate the residue of the ##f(z)=z^2 \sin \dfrac {1}{z+1}## at its singular points.
I'm not sure what should I do. Should I expand both ##z^2## and ##\sin \dfrac {1}{z+1}## at ##z=-1##?

Any help would be appreciated.

1. The sine expansion should have ##(z-\pi)^{2n+1}##, not ##(z-\pi)^{2n}##

2. No. The sine has all odd ##z## in the expansion so in particular the ##z^3## term from the sine becomes a 1/z term when divided by ##z^4##.

MatinSAR
Orodruin said:
1. The sine expansion should have ##(z-\pi)^{2n+1}##, not ##(z-\pi)^{2n}##
It's clear now.
Orodruin said:
2. No. The sine has all odd ##z## in the expansion so in particular the ##z^3## term from the sine becomes a 1/z term when divided by ##z^4##.
There should be a mistake in my expansion. I will try again.

Thank you for your time.

Orodruin said:
1. The sine expansion should have ##(z-\pi)^{2n+1}##, not ##(z-\pi)^{2n}##

2. No. The sine has all odd ##z## in the expansion so in particular the ##z^3## term from the sine becomes a 1/z term when divided by ##z^4##.
$$f(z) = \dfrac{\sin(z)}{z^4} = \dfrac{z - \dfrac{z^3}{3!} + \dfrac{z^5}{5!} - \dfrac{z^7}{7!} + \cdots}{z^4}$$Therefore, the residue of ## f(z) = \frac{\sin(z)}{z^4} ## at ## z = 0## is ## -\frac{1}{6}##. Thank you for your help.

Last edited:
Orodruin

### Similar threads

• Calculus and Beyond Homework Help
Replies
3
Views
627
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
13
Views
1K
• Calculus and Beyond Homework Help
Replies
11
Views
545
• Calculus and Beyond Homework Help
Replies
3
Views
746
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
689
• Calculus and Beyond Homework Help
Replies
6
Views
4K
• Calculus and Beyond Homework Help
Replies
5
Views
213