# Residue Theorem: Evaluating Integral |z|=1 (sin(z)/z^2)dz

• brianhawaiian
In summary, the student is trying to solve an integral but is having trouble because z = 0 is an essential singularity. The student uses the residue theorem to find the residue at z = 0.
brianhawaiian
Sorry I don't have equation editor, for some reason every time I install it on Microsoft Word it never appears...

## Homework Statement

Calculate the residue at each isolated singularity in the complex plane

e^(1/z)

## Homework Equations

#1 Simple pole at z0 then,
Res[f(z), z0] = lim (z - z0)(f(z)) as z goes to z0.
#2 Double pole at z0 then,
Res[f(z), z0] = lim d/dz [(z - z0)^2*f(z)] as z goes to z0.
#3 If f(z) and g(z) are analytic at z0, and if g(z) has a simple zero at z0 then
Res[f(z)/g(z), z0] = f(z0)/g'(z0)
#4 If g(z) is analytic and has a simple zero at z0, then
Res[1/g(z), z0] = 1/g'(z0).

## The Attempt at a Solution

The problem occurs when z = 0 so looking at

Res[e^(1/z), 0], Using #1, #3, #4 don't help the problem. So using #2

lim as z goes to z0 [d/dz z^2 * e^(1/z)], there's still a problem... I'm completely lost at this point.

## Homework Statement

Evaluate the following integral, using the residue theorem
Integral |z| = 1 (sin(z)/z^2)dz

See Above

## The Attempt at a Solution

How would I start this? z = 0 gives a problem, would I take the integral first and then evaluate?

z=0 isn't a pole for this function.

write out the laurent series. it has all negative powers. thus z=0 is an essential isolated singularity.

knowing the laurent series, how do you find the residue?

New Edit: I agree with Latentcorpse (he replied whe I was still typing the message)

You know that exp(z) is an analytical function everywhere on the complex plane. This means that you can replace z by 1/z in the series expansion without any problem. It will yield a Laurent series that converges everywhere (except at z = 0). You can then read-off the residue at z = 0.

The Laurent expansion contains arbitrarily large negative powers of z, we then say that the function has an "essential singularity" at z = 0.

You can still compute the residue at this essential singularity like in case of ordinary poles, by considering the mapping z ---> 1/z

The residue is 1/(2 pi i) times a contour integral that encircles the singularity in an anti-clockwise way. If then perform a change of variables z -->1/z, the integral changes to:

1/(2 pi i) exp(z) (-1/z^2) dz

which is now traversed clockwise. Changing it to anticlockwise will yield a minus sign. This means that the residue at zero of exp(1/z) is the same as the residue at zero of the function

exp(z)/z^2

This function has an ordinary pole (a "double pole") at zero.

Thank you, appreciate the comments... I'm completely lost in Complex Analysis (First Grad Level class, never took it as an undergrad)

read off the $c_{-1}$ coefficient i.e. the coefficient of the $z^{-1}$ term in the laurent series for $\frac{e^z}{z^2}$ and you'll get the residue is 1. hopefully.

latentcorpse said:
read off the $c_{-1}$ coefficient i.e. the coefficient of the $z^{-1}$ term in the laurent series for $\frac{e^z}{z^2}$ and you'll get the residue is 1. hopefully.

Yeah, I figured it out thanks to you guys, much appreciated, I have another question up on the page, was wondering if you could maybe help with that one? If not it's okay, Thanks again!

## What is the Residue Theorem?

The Residue Theorem is a mathematical tool used in complex analysis to evaluate integrals around closed curves, also known as contours. It states that the value of an integral around a closed contour is equal to the sum of the residues of the function inside the contour.

## What is the significance of "Residue Theorem: Evaluating Integral |z|=1 (sin(z)/z^2)dz"?

This specific example of the Residue Theorem is used to evaluate integrals of the form ∫(f(z)/z^m)dz, where m is a positive integer. It is particularly useful in calculating integrals involving trigonometric functions, such as in this case where the function is sin(z).

## What is a residue?

In complex analysis, a residue is the value of a function at a point where it is not defined. It is calculated by taking the limit of the function as it approaches the point. In the context of the Residue Theorem, it is the coefficient of the (z-a)^-1 term in the Laurent series expansion of the function, where a is the point inside the contour.

## How do you find the residues for this integral?

To find the residues for this integral, we first need to find the Laurent series expansion of the function sin(z)/z^2. This can be done by using the Maclaurin series expansion of sin(z) and then dividing by z^2. The coefficients of the (z-a)^-1 term in this expansion will be the residues. In this case, there is only one pole at z=0, so the residue can be found by simply evaluating the limit of the function as z approaches 0.

## What are the applications of the Residue Theorem?

The Residue Theorem has many applications in mathematics and physics. It is commonly used in contour integration to solve complex integrals, and is also used in solving differential equations, finding singularities of functions, and calculating complex line integrals. It has also been applied in fields such as fluid mechanics, quantum mechanics, and signal processing.

• Calculus and Beyond Homework Help
Replies
2
Views
946
• Calculus and Beyond Homework Help
Replies
14
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
907
• Engineering and Comp Sci Homework Help
Replies
6
Views
928
• Calculus and Beyond Homework Help
Replies
13
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
676
• Engineering and Comp Sci Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
617
• Calculus and Beyond Homework Help
Replies
6
Views
825
• Calculus and Beyond Homework Help
Replies
2
Views
557