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

[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

Homework Equations

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?

 

[latentcorpse]

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?

 

[Count Iblis]

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.

 

[brianhawaiian]

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

 

[latentcorpse]

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.

 

[brianhawaiian]

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!