Residue theory can this be right

[tilika123]

if f(z) = e^z/(z^2 - 1)

does Res[f,-1] = -1/2e
and Res[f,1] = e/2

 

[shmoe]

What makes you think those are wrong?

 

[tilika123]

im just learning the theory and i did have the concept wrong until you helped me now i just want to make sure i can use it correctly. If you would like to give me a problem to solve that would be great. thanks

 

[shmoe]

Have some confidence! Finding the residues at simple poles like these is no more complicated than finding a limit, which you've had lots of practice doing.

You had these two correct, hopefully your book has more practice? Here's a couple problems if you like, find the poles and calculate residues:

\frac{\sin z}{z^2}

\frac{1}{\sin \pi z}

\frac{e^z}{z^2-z}

Have you talked about poles of higher order than 1 yet?

 

[tilika123]

yes we have looked at order of 1 and 2 and k. My understanding of order is if there is one zero there is one pole, 2 zeros 2 poles, if there are zeros that ore the same then the order goes up. Example 1/(z-1)^6 has one pole order of 6. we breifly discussed laurent series but not enough. we basically saw a example of it and the book we have is not real good.

Res[sinz/z^2,0] = 1 has 1 pole order 2

as for the second there could be a pole at z = 1 or z = 0 so would i calculate residues for both

3 Res[e^z/(z^2-z),1]= e
and Res[e^z/(z^2-z),0]= -1

 

[shmoe]

yes we have looked at order of 1 and 2 and k. My understanding of order is if there is one zero there is one pole, 2 zeros 2 poles, if there are zeros that ore the same then the order goes up. Example 1/(z-1)^6 has one pole order of 6. we breifly discussed laurent series but not enough. we basically saw a example of it and the book we have is not real good.

That works. Higher order poles will mean the Laurent series has higher powers of (z-z0)^(-1) in it, if f(z) has a pole of order 2 at z0, then we'd have

f(z)=\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{(z-z_0)}+a_{0}+a_1 (z-z_0)+\ldots

where a_{-2} is non zero. Again the residue is a_{-1}. You should think about how you would calculate this residue and how it relates to this Laurent series.

Laurent series are nothing to be afraid of, they are just power series with negative exponents, but converge in annuli instead of discs.

Res[sinz/z^2,0] = 1 has 1 pole order 2

But the numerator has a zero of order 1 at z=1, so sinz/z^2 will have a pole of order 1 only at z=0. It's a fluke that treating it as a pole of order 2 ends up with the corect residue.

as for the second there could be a pole at z = 1 or z = 0 so would i calculate residues for both

there's more! sin pi*z has a zero at every integer...but it's periodic so it's enough to do it for z=0 and z=1. Can you find these residues?

3 Res[e^z/(z^2-z),1]= e
and Res[e^z/(z^2-z),0]= -1

good!

 

[tilika123]

That works. Higher order poles will mean the Laurent series has higher powers of (z-z0)^(-1) in it, if f(z) has a pole of order 2 at z0, then we'd have

f(z)=\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{(z-z_0)}+a_{0}+a_1 (z-z_0)+\ldots

where a_{-2} is non zero. Again the residue is a_{-1}. You should think about how you would calculate this residue and how it relates to this Laurent series.

Laurent series are nothing to be afraid of, they are just power series with negative exponents, but converge in annuli instead of discs.




But the numerator has a zero of order 1 at z=1, so sinz/z^2 will have a pole of order 1 only at z=0. It's a fluke that treating it as a pole of order 2 ends up with the corect residue.

I don't get it how does sinz have a zero at z = 1

there's more! sin pi*z has a zero at every integer...but it's periodic so it's enough to do it for z=0 and z=1. Can you find these residues?

no i can't with the formulas i have there will be a zero in the denominator

when Res[1/zsin(piz),0] = lim z->0 (z-0)f(x)
=lim z -> 0 of 1/sin(0) undefined

 

[shmoe]

I don't get it how does sinz have a zero at z = 1

It doesn't! And don't listen to the guy who told you it does! Sorry, I meant sin(z) has a zero at z=0, which 'cancels' with one of the zeros of the z^2 in the denominator

no i can't with the formulas i have there will be a zero in the denominator
when Res[1/zsin(piz),0] = lim z->0 (z-0)f(x)
=lim z -> 0 of 1/sin(0) undefined

You have an extra 1/z in there:

Res\left[\frac{1}{\sin \pi z},0\right]=\lim_{z\rightarrow 0}\frac{z}{\sin \pi z}=\frac{1}{\pi}\lim_{z\rightarrow 0}\frac{\pi z}{\sin \pi z}=\frac{1}{\pi}

To find the residue at 1, you can write \sin \pi z=-\sin \pi (z-1) and it's similar to the above.

 

[tilika123]

i still don't get it as z approaches 0 sin piz goes to 0
How do you pull 1/pi out?

 

[shmoe]

It's your old friend from first year calculus,

\lim_{x\rightarrow 0}\frac{\sin x}{x}=1

 

[tilika123]

thanks again
I have one more question for you if you have a function that has poles of say 3 and -3 order 1 and a singularity in the numerator of 0 what would happen. i guess an example would be z/(z^2-9)

 

[shmoe]

It still has poles of order 1 at z=3 and z=-3 and a zero of order 1 at z=0. This is for your z/(z^2-9) whose numerator has a zero, not a singularity at z=0.

If you have f(z)=g(z)/h(z) and g has a zero of order m at z0 while h has a zero of order n at z0. If m-n<0, then f has a pole of order m-n at z0. If m-n>=0 and n>0 we have a removable singularity, and will end up with a zero of order m-n.

e.g. f(z)=\frac{(\sin z)^2}{z}

then f(z) has a removable singularity at z=0, and we can define f(0)=0 to get an analytic function. f will then have a zero of order 1 at z=0.

1/f(z) will have a pole of order 1 at z=0, and poles of order 2 at all the other zeros of sin z

 

[tilika123]

Yes but what would happen when the poles differ from the zeros in the numerator in the example i gave z/(z^2-9) in this case would m-n = 0 or can you not do that because is has poles of 3.-3 and a zero 0f 0.

 

[shmoe]

Yes but what would happen when the poles differ from the zeros in the numerator in the example i gave z/(z^2-9) in this case would m-n = 0 or can you not do that because is has poles of 3.-3 and a zero 0f 0.

You treat each zero seperately. z=-3, 3, and 0 will all have their own values of "m" and "n" for z/(z^2-9):

at z=3, the numerator has no zero, so m=0. The denominator has a zero of order 1, so n=1 and m-n=-1, and hence a pole of order n-m=1 (I made a typo in my last post for this case, if m-n<0, the pole is of order n-m, should be clear from examples)

same thing at z=-3, get a pole of order 1

at z=0, the numerator has a zero of order 1, so m=1. the denominator has no zero, so n=0, and hence a zero of order m-n=1-0=1.

You can find a batch of residue (and other) problems of various difficulties with solutions here
http://www.exampleproblems.com/wiki/index.php/Complex_Variables#Residues

 

[tilika123]

thank you for your help and time