Residue Theorem applied to a keyhole contour

In summary, the conversation is about solving an integral using the residue theorem. The first step is to factorize the denominator and find values for z0 that correspond to isolated singularities. The next step is to calculate the residue at each simple pole, which is done by multiplying the function by (z-z0) and evaluating it at z0. The final step is to use the residue theorem, which states that the integral is equivalent to 2*pi*i times the sum of the residues. The conversation also covers finding the values of z0 in polar form, and correcting errors in the calculations.
  • #1
16
3
Homework Statement
Use the Residue Theorem to show:

$$\int \:\frac{z^{\frac{1}{2}}}{1+\sqrt{2}z+z^2}dz=2^{\frac{2}{3}}\pi isin\left(\frac{8\pi }{3}\right)$$

for a keyhole contour where ##z=re^{i\theta }## and ##-\pi <\theta <\pi##
Relevant Equations
the standard residue theorem
I'm really struggling with this one. A newbie to using the residue theorem. I'm trying to solve this by factorising the denominator to find values for z0 and I have:

##z=\frac{-\sqrt{2}+i\sqrt{2}}{2}## and ##z=\frac{-\sqrt{2}-i\sqrt{2}}{2}##

I also know that sin(3π/8)= ##\frac{\sqrt{2+\sqrt{2}}}{2}##
 
Last edited:
Physics news on Phys.org
  • #2
Show us residues you calculated, please.
 
  • Like
Likes topsquark
  • #3
The path of integration is not clear to me. I can make a guess, but it is better if you spell it out. What is the multiplier 'a'? What is a "keyhole contour"?
 
  • #4
Hello,
Thanks for the responses. The integral is a closed contour integral (with a branch cut along the negative x-axis so the function is not multi-valued) but I couldn't find the correct symbol with the integral sign having a circle. The residue theorem states that in place of calculating the integral explicitly, it is equivalent to 2*pi*i*sum of residues.
 
  • #5
anuttarasammyak said:
Show us residues you calculated, please.
I haven't that's the problem. The first step, in the process I'm familiar with, is to factorise the denominator so that you can find values of z for which their are isolated singularities. I've tried to do that and I've given my results in the original post. Not sure where to go next..

p.s. I'm aware you can use the Laurent series, but my series expansion knowledge isn't great. Hence I'm trying to compute this using poles.
 
Last edited:
  • #6
FactChecker said:
The path of integration is not clear to me. I can make a guess, but it is better if you spell it out. What is the multiplier 'a'? What is a "keyhole contour"?
the "a" was a mistake, sorry. I've removed it
 
  • #7
FactChecker said:
What is a "keyhole contour"?
This is an example:

contour.jpg
 
  • Like
Likes DrClaude
  • #8
Teymur said:
I haven't that's the problem. The first step, in the process I'm familiar with, is to factorise the denominator so that you can find values of z for which their are isolated singularities. I've tried to do that and I've given my results in the original post. Not sure where to go next..
These are simple poles, so what's the procedure for calculating the residue of a simple pole?
 
  • #9
Teymur said:
$$\int \:\frac{z^{\frac{1}{2}}}{1+\sqrt{2}z+z^2}dz=2^{\frac{2}{3}}\pi isin\left(\frac{8\pi }{3}\right)$$
Please correct the typos in your work. There's a fairly obvious one, but I suspect there are more.
 
  • #10
vela said:
Please correct the typos in your work. There's a fairly obvious one, but I suspect there are more.
Ah yes, so sorry!! The integral should give:

$$2^{\frac{3}{2}}\pi \:isin\left(\frac{3\pi }{8}\right)$$

I got the numerators and denominators mixed up! Soz
 
  • #11
vela said:
These are simple poles, so what's the procedure for calculating the residue of a simple pole?
Yes exactly, simple poles. Firstly, I'm just checking to see if the values I have obtained for z0 at the poles are correct, they seem unwieldly. If they are correct, you multiply f(z) by (z-z0) and evaluate the result at =z0.. I've tried to do that but can't see how it would get you to the intended outcome.
 
  • #12
Teymur said:
Yes exactly, simple poles. Firstly, I'm just checking to see if the values I have obtained for z0 at the poles are correct, they seem unwieldly.
The values of ##z_0## seem right. What are they in polar form?
 
  • Like
Likes vela
  • #13
Teymur said:
Firstly, I'm just checking to see if the values I have obtained for z0 at the poles are correct, they seem unwieldly.
You can always check them by plugging them back into the quadratic and seeing if the result is 0. They are indeed a bit unwieldy for some parts of the calculation. As @julian suggested, you'll find it convenient to express them in polar form.

Teymur said:
If they are correct, you multiply f(z) by (z-z0) and evaluate the result at =z0. I've tried to do that but can't see how it would get you to the intended outcome.
Post what you get. Remember you have two poles, so it's the sum that will produce the result you want.
 
  • Like
Likes julian and Teymur
  • #14
Okay cool. Am I right that you get:

$$\cos \left(\frac{\pi }{4}\right)+isin\left(\frac{\pi }{4}\right)=e^{\frac{i\pi }{4}}$$

and

$$\cos \left(\frac{3\pi i}{4}\right)+isin\left(\frac{3\pi i}{4}\right)=e^{\frac{3\pi i}{4}}$$

And if so, is the next step:

$$\left(z-e^{\frac{\pi i}{4}}\right)f\left(z\right)=\left(z-e^{\frac{\pi i}{4}}\right)\frac{z^{\frac{1}{2}}}{\left(z-e^{\frac{\pi i}{4}}\right)\left(z-e^{\frac{3\pi i}{4}}\right)}$$

For ##z=e^{\frac{\pi i}{4}}##

$$\frac{e^{\frac{\pi i}{8}}}{\left(e^{\frac{\pi i}{4}}-e^{\frac{3\pi i}{4}}\right)}$$
 
Last edited:
  • #15
The first one isn't correct, but the second one is. (Just check the signs.) Note that because the coefficients in the quadratic are real, the roots will be complex conjugates of each other. So this means the other root is ##e^{-3\pi i/4}##.

The next step will look okay once you fix the roots. I'd note it's a limit as ##z \to z_0##.
 
  • Like
Likes julian
  • #16
vela said:
The first one isn't correct, but the second one is. (Just check the signs.) Note that because the coefficients in the quadratic are real, the roots will be complex conjugates of each other. So this means the other root is ##e^{-3\pi i/4}##.

The next step will look okay once you fix the roots. I'd note it's a limit as ##z \to z_0##.
You mean that the roots are ##e^{\frac{3\pi i}{4}}## and ##e^{\frac{-3\pi i}{4}}##?
 
  • Like
Likes julian
  • #17
Are you saying:

$$\left(z-e^{\frac{3\pi i}{4}}\right)f\left(z\right)=\left(z-e^{\frac{3\pi i}{4}}\right)\frac{z^{\frac{1}{2}}}{\left(z-e^{\frac{3\pi i}{4}}\right)\left(z-e^{\frac{-3\pi i}{4}}\right)}$$

so that as z goes to z0:

$$\frac{e^{\frac{3\pi i}{8}}}{\left(e^{\frac{3\pi i}{4}}-e^{\frac{-3\pi i}{4}}\right)}$$
 
  • #18
Teymur said:
Are you saying:

$$\left(z-e^{\frac{3\pi i}{4}}\right)f\left(z\right)=\left(z-e^{\frac{3\pi i}{4}}\right)\frac{z^{\frac{1}{2}}}{\left(z-e^{\frac{3\pi i}{4}}\right)\left(z-e^{\frac{-3\pi i}{4}}\right)}$$

so that as z goes to z0:

$$\frac{e^{\frac{3\pi i}{8}}}{\left(e^{\frac{3\pi i}{4}}-e^{\frac{-3\pi i}{4}}\right)}$$
Do the same thing with the root ##e^{-\frac{3\pi i}{4}}##. Add the two residues, then multiply by ##2 \pi i##.
 
Last edited:
  • Like
Likes Teymur
  • #19
julian said:
Do the same thing with the root ##e^{-\frac{3\pi i}{4}}##. Add the two residues, then multiply by ##2 \pi i##.
working backwrads from the solution given in the question and putting aside the prefactor of ##2\pi i## that comes from the theorem, that means that the sum of the residues should give ##2^{\frac{1}{2}}sin\left(\frac{3\pi }{8}\right)##

I can't see how to get there from:

$$\frac{e^{\frac{3\pi \:i}{8}}}{e^{\frac{3\pi \:i}{4}}-e^{-\frac{3\pi \:i}{4}}}+\frac{e^{\frac{-3\pi \:i}{8}}}{e^{\frac{-3\pi \:i}{4}}-e^{\frac{3\pi \:i}{4}}}$$
 
  • #20
I'm assuming the following identity has something to do with it, but can't seem to put it all together:

$$sin\left(x\right)=\frac{e^{ix}-e^{-ix}}{2i}$$
 
  • Like
Likes FactChecker
  • #21
Teymur said:
working backwrads from the solution given in the question and putting aside the prefactor of ##2\pi i## that comes from the theorem, that means that the sum of the residues should give ##2^{\frac{1}{2}}sin\left(\frac{3\pi }{8}\right)##

I can't see how to get there from:

$$\frac{e^{\frac{3\pi \:i}{8}}}{e^{\frac{3\pi \:i}{4}}-e^{-\frac{3\pi \:i}{4}}}+\frac{e^{\frac{-3\pi \:i}{8}}}{e^{\frac{-3\pi \:i}{4}}-e^{\frac{3\pi \:i}{4}}}$$
The denominator of the 2nd term is the negative of the denominator of the first term. That allows you to write your expression as a single fraction.

And then, yes, you use ##\sin (x) = \frac{e^{i x} - e^{-i x}}{2i}##.
 
Last edited:
  • Like
Likes FactChecker
  • #22
I get the feeling this isn't the best way to simplify but what I get is as follows:

$$\frac{sin\left(\frac{3\pi }{8}\right)}{sin\left(\frac{3\pi }{4}\right)}=\sqrt{\frac{2+\sqrt{2}}{2}}$$

If you then multiply through by ##2^{\frac{1}{2}}## you get:

$$2^{\frac{1}{2}}\frac{\sqrt{2+\sqrt{2}}}{2}=2^{\frac{1}{2}}sin\left(\frac{3\pi \:}{8}\right)$$

as required. If there's a simpler, straightforward way to do this, I'd be grateful to hear.
 
  • #23
The sum of residues is indeed equal to:

\begin{align*}
\dfrac{\sin \left( \frac{3 \pi}{8} \right)}{\sin \left( \frac{3 \pi}{4} \right)}
\end{align*}

You then just substitute ##\sin \left( \frac{3 \pi}{4} \right) = \frac{1}{\sqrt{2}}## into it to obtain the desired result.
 
Last edited:
  • Like
Likes FactChecker, Teymur and SammyS
  • #24
Appreciate all the support
 
  • Like
Likes julian

Suggested for: Residue Theorem applied to a keyhole contour

Replies
10
Views
605
Replies
2
Views
767
Replies
1
Views
423
Replies
3
Views
516
Replies
12
Views
1K
Replies
3
Views
321
Replies
14
Views
1K
Replies
26
Views
1K
Back
Top