Contour Integral and Cauchy's residue theorem

[Ted123]

I've tried this using the definition of a contour integral and Cauchy's residue theorem but get conflicting answers.

\displaystyle f(z) = \frac{z+1}{z(2z-1)(z+2)}

We can parametrise the contour \gamma (the unit circle) by \gamma(t) = e^{it} for t \in [0, 2\pi ]

So by the definition of a contour integral

2\pi i \times I = \displaystyle \int^{2\pi}_0 \frac{e^{it}+1}{2e^{3it} + 3e^{2it} - 2e^{it}} \times ie^{it}\;dt = \frac{\pi i}{5}

But using the residue therem

\displaystyle \text{res}(f,0) = -\frac{1}{2}

\displaystyle \text{res} \left( f,\frac{1}{2} \right) = \frac{6}{5}

\displaystyle \text{res} (f,-2) = -\frac{1}{10}

\gamma wraps once (in the negative direction) around z=0 and z=\frac{1}{2} so n (\gamma, 0) = n (\gamma, \frac{1}{2}) = -1 and n (\gamma, -2) = 0

So \displaystyle 2\pi i \times I = 2\pi i \left(-\frac{1}{2} \times -1 + \frac{6}{5} \times -1 \right) = -\frac{7\pi i}{5}

which is not what I got before.

 

[Dick]

I've tried this using the definition of a contour integral and Cauchy's residue theorem but get conflicting answers.

\displaystyle f(z) = \frac{z+1}{z(2z-1)(z+2)}

We can parametrise the contour \gamma (the unit circle) by \gamma(t) = e^{it} for t \in [0, 2\pi ]

So by the definition of a contour integral

2\pi i \times I = \displaystyle \int^{2\pi}_0 \frac{e^{it}+1}{2e^{3it} + 3e^{2it} - 2e^{it}} \times ie^{it}\;dt = \frac{\pi i}{5}

But using the residue therem

\displaystyle \text{res}(f,0) = -\frac{1}{2}

\displaystyle \text{res} \left( f,\frac{1}{2} \right) = \frac{6}{5}

\displaystyle \text{res} (f,-2) = -\frac{1}{10}

\gamma wraps once (in the negative direction) around z=0 and z=\frac{1}{2} so n (\gamma, 0) = n (\gamma, \frac{1}{2}) = -1 and n (\gamma, -2) = 0

So \displaystyle 2\pi i \times I = 2\pi i \left(-\frac{1}{2} \times -1 + \frac{6}{5} \times -1 \right) = -\frac{7\pi i}{5}

which is not what I got before.

You are going around counterclockwise. I don't think that makes the winding number -1. And Res(f,1/2) isn't 6/5. Can you show how you got that?

 

[tylerc1991]

You have to use the residue theorem in this case because of the singularity at z = 0. Assuming the residues were calculated correctly, the second answer is correct.

 

[Dick]

You have to use the residue theorem in this case because of the singularity at z = 0. Assuming the residues were calculated correctly, the second answer is correct.

You don't HAVE to use residues. The answer in the first method is correct. Though I don't know how it was done. Probably by typing into Wolfram Alpha or something. And the residues aren't correct. That's the big problem. Sometimes I think it should be a rule here that people actually work a problem out before feeling qualified to give advice.

 

[tylerc1991]

You don't HAVE to use residues. The answer in the first method is correct. Though I don't know how it was done. Probably by typing into Wolfram Alpha or something. And the residues aren't correct. That's the big problem. Sometimes I think it should be a rule here that people actually work a problem out before feeling qualified to give advice.

I completely agree with you, and it was a hasty comment to make on my behalf. On the other hand, seeing as how the number of people who answer questions seem to be fairly outnumbered (at least from what I have seen), the process would completely clog up if it were required to work a problem out before commenting. While one person takes the time to work a problem out, 5 more have been posted, etc. Often times, people just need a slight push in the right direction, which doesn't involve someone working the entire problem. This is how group brainstorming works. Of course this doesn't apply in this case because my comment wasn't helpful, but the principle still holds. That said, I will keep your comment in mind next time I think about commenting so as to avoid ridicule.

 

[Dick]

I completely agree with you, and it was a hasty comment to make on my behalf. On the other hand, seeing as how the number of people who answer questions seem to be fairly outnumbered (at least from what I have seen), the process would completely clog up if it were required to work a problem out before commenting. While one person takes the time to work a problem out, 5 more have been posted, etc. Often times, people just need a slight push in the right direction, which doesn't involve someone working the entire problem. This is how group brainstorming works. Of course this doesn't apply in this case because my comment wasn't helpful, but the principle still holds. That said, I will keep your comment in mind next time I think about commenting so as to avoid ridicule.

Oh, I didn't mean to ridicule. Sorry. But still, there is a fairly specific problem here that generalized 'group brainstorming' isn't going to catch.

 

[Ted123]

res(f,1/2) should be 3/5 and the winding numbers should be 1 (anticlockwise means 'positively oriented'), so I think this is now correct:

\displaystyle \text{res}(f,0) = -\frac{1}{2}

\displaystyle \text{res} \left( f,\frac{1}{2} \right) = \frac{3}{5}

\displaystyle \text{res} (f,-2) = -\frac{1}{10}

\gamma wraps once (in the negative direction) around z=0 and z=\frac{1}{2} so n (\gamma, 0) = n (\gamma, \frac{1}{2}) = 1 and n (\gamma, -2) = 0

So \displaystyle 2\pi i \times I = 2\pi i \left(-\frac{1}{2} \times 1 + \frac{3}{5} \times 1 \right) = \frac{\pi i}{5}

So I = \frac{1}{10}

I typed the first integral into Wolfram Alpha merely to check whether the 2 answers agreed, which they didn't!