# Residue Theorem applied to a keyhole contour

Teymur
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}##

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Gold Member
Show us residues you calculated, please.

• topsquark
Homework Helper
Gold Member
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"?

Teymur
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.

Teymur
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.

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Teymur
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

Gold Member
What is a "keyhole contour"?
This is an example: • DrClaude
Staff Emeritus
Homework Helper
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?

Staff Emeritus
Homework Helper
$$\int \:\frac{z^{\frac{1}{2}}}{1+\sqrt{2}z+z^2}dz=2^{\frac{2}{3}}\pi isin\left(\frac{8\pi }{3}\right)$$
Please correct the typos in your work. There's a fairly obvious one, but I suspect there are more.

Teymur
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

Teymur
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.

Gold Member
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?

• vela
Staff Emeritus
Homework Helper
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.

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.

• julian and Teymur
Teymur
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}}##

$$\frac{e^{\frac{\pi i}{8}}}{\left(e^{\frac{\pi i}{4}}-e^{\frac{3\pi i}{4}}\right)}$$

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Staff Emeritus
Homework Helper
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##.

• julian
Teymur
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}}##?

• julian
Teymur
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:

$$\frac{e^{\frac{3\pi i}{8}}}{\left(e^{\frac{3\pi i}{4}}-e^{\frac{-3\pi i}{4}}\right)}$$

Gold Member
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##.

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• Teymur
Teymur
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:

$$\frac{e^{\frac{3\pi \:i}{8}}}{e^{\frac{3\pi \:i}{4}}-e^{-\frac{3\pi \:i}{4}}}+\frac{e^{\frac{-3\pi \:i}{8}}}{e^{\frac{-3\pi \:i}{4}}-e^{\frac{3\pi \:i}{4}}}$$

Teymur
I'm assuming the following identity has something to do with it, but can't seem to put it all together:

$$sin\left(x\right)=\frac{e^{ix}-e^{-ix}}{2i}$$

• FactChecker
Gold Member
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}##.

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• FactChecker
Teymur
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.

Gold Member
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.

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• FactChecker, Teymur and SammyS
Teymur
Appreciate all the support

• julian