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    First time I tried it was by deforming the square into a circle of radius 2 with centre 0, however this does not seem to provide the right answer. I tried it a second time by deforming it into a circle of radius 1, center -1, and it worked out wonders.

    So why/where did it go wrong in the first attempt? The singularity is still strictly inside the circle of radius 2, and the function \dfrac{z}{z+1} is analytic between the circle and the square, as well as on both.

    Is there something else I'm missing? I think I'm probably going wrong somewhere with the Log part but I'm not sure.


    Deformed contour would be C' : z=2e^{i \theta} for \theta \in [0, 2\pi)

    So
    \begin{aligned} \int_C \frac{z}{z+1} .dz & = \int_{C'} (1-\frac{1}{z+1}) .dz \\ & = \int_{0}^{2\pi} (1-\frac{1}{2e^{i\theta}+1}) \cdot 2ie^{i \theta} .d\theta \\ & = \int_0^{2\pi}2ie^{i \theta} - \frac{2ie^{i\theta}}{2e^{i\theta  }+1} .d \theta \\ & =2 \underbrace{\int_0^{2\pi} ie^{i \theta}}_{=0} + \mathrm{Log}[2e^{i\theta}+1]_0^{2\pi} \\ & = \log(3) - \log 3 \\ & = 0 \\ & \neq -2\pi i
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    (Original post by RDKGames)
    ...
    'Fraid my complex analysis is non-existant.
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    (Original post by ghostwalker)
    'Fraid my complex analysis is non-existant.
    That makes two of us then
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    I would think a bit more carefully about the branch of the logarithm you're using there.
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    (Original post by Gregorius)
    I would think a bit more carefully about the branch of the logarithm you're using there.
    This is what I can't seem to get my head around for some reason.

    I defined \theta \in [0,2\pi) so the principal arguments are in this region, hence I'm not sure how I can get a 2\pi or any multiple of it, when it comes to evaluating the log of the complex number, as they would just go to the principal argument of 0.

    Unless I say something like \log(2e^{2\pi i}+1) = \log 3 and \log (2e^{0\pi i}+1) = \log 3 + 2\pi i but TBH I'm confused enough to not see where this would come from or why what I've written is not valid.
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    (Original post by RDKGames)
    \displaystyle \int_0^{2\pi}2ie^{i \theta} - \dfrac{2ie^{i\theta}}{2e^{i\thet  a}+1} d \theta  =2 \underbrace{\int_0^{2\pi} ie^{i \theta} d\theta}_{=0} + \mathrm{Log}[2e^{i\theta}+1]_0^{2\pi} = \log 3 - \log 3
    First thing to note is that there is a sign error here in the former equality, the relevance of which will become obvious in a moment. This second is that, by virtue of using the complex logarithm, you've made a choice of branch and consequently cannot expect both terms to evaluate to \log 3 as the two \thetas differ by a multiple of 2\pi.

    (Original post by RDKGames)
    This is what I can't seem to get my head around for some reason.

    I defined \theta \in [0,2\pi) so the principal arguments are in this region, hence I'm not sure how I can get a 2\pi or any multiple of it, when it comes to evaluating the log of the complex number, as they would just go to the principal argument of 0.

    Unless I say something like \log(2e^{2\pi i}+1) = \log 3 and \log (2e^{0\pi i}+1) = \log 3 + 2\pi i but TBH I'm confused enough to not see where this would come from or why what I've written is not valid.
    You're right in saying that you need something like the above, and the barrier appears to be arising from a misunderstanding of the principle value of the logarithm. Recall that the p.v. of the logarithm (with your convention of [0,2π) defining "principle" ) is:

    \text{Log } z = \log |z| + i\text{Arg }z, where 0\leq \text{Arg } z < 2\pi.

    This is defined by a branch cut along the positive real axis, and from this definition it should be obvious that:

    \mathrm{Log}[2e^{i\theta}+1]_0^{2\pi} = (\log 3 +2\pi i) - (\log 3 + 0\pi i) = 2\pi i \not= 0

    After you correct the sign error above, this is then consistent with the result you obtained using the other circular contour.

    One slightly subtle point about the arguments - as a consequence of the branch we've chosen to work with, if z_+ is 'just above' the positive real axis and z_- 'just below' the positive real axis, we have \text{Arg }(z_+) = 0, \text{Arg }(z_-) = 2\pi. If you're confused by the fact that \text{Arg }(z_-) is not in the interval of principle arguments, think about what happens to the argument as \theta approaches 2\pi in the limit - in order to return to an argument of zero, we would have to cross over a discontinuity of the log and this can't happen.

    Notice that, when \theta=\theta_+ is 'just above' \ 0, z_+=2e^{i\theta_+} +1 is 'close' to z=3 and 'just above' the real axis i.e. \arg{z_+} = 0. On the other hand, when \theta=\theta_- is 'just below' 2\pi, z_- = 2e^{i\theta_-} + 1 is still close to z=3 but 'just below' it i.e. \arg{z_-} = 2\pi.
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    (Original post by Farhan.Hanif93)
    First thing to note is that there is a sign error here in the former equality, the relevance of which will become obvious in a moment. This second is that, by virtue of using the complex logarithm, you've made a choice of branch and consequently cannot expect both terms to evaluate to \log 3 as the two \thetas differ by a multiple of 2\pi.


    You're right in saying that you need something like the above, and the barrier appears to be arising from a misunderstanding of the principle value of the logarithm. Recall that the p.v. of the logarithm (with your convention of [0,2π) defining "principle" ) is:

    \text{Log } z = \log |z| + i\text{Arg }z, where 0\leq \text{Arg } z < 2\pi.

    This is defined by a branch cut along the positive real axis, and from this definition it should be obvious that:

    \mathrm{Log}[2e^{i\theta}+1]_0^{2\pi} = (\log 3 +2\pi i) - (\log 3 + 0\pi i) = 2\pi i \not= 0

    After you correct the sign error above, this is then consistent with the result you obtained using the other circular contour.

    One slightly subtle point about the arguments - as a consequence of the branch we've chosen to work with, if z_+ is 'just above' the positive real axis and z_- 'just below' the positive real axis, we have \text{Arg }(z_+) = 0, \text{Arg }(z_-) = 2\pi. If you're confused by the fact that \text{Arg }(z_-) is not in the interval of principle arguments, think about what happens to the argument as \theta approaches 2\pi in the limit - in order to return to an argument of zero, we would have to cross over a discontinuity of the log and this can't happen.

    Notice that, when \theta=\theta_+ is 'just above' \ 0, z_+=2e^{i\theta_+} +1 is 'close' to z=3 and 'just above' the real axis i.e. \arg{z_+} = 0. On the other hand, when \theta=\theta_- is 'just below' 2\pi, z_- = 2e^{i\theta_-} + 1 is still close to z=3 but 'just below' it i.e. \arg{z_-} = 2\pi.
    Ah this clears a lot of things up! I hadn't considered taking the approach of splitting theta's into 'from below' and 'from above' and it makes sense now why we allow the 2\pi

    As for the sign error, I should've seen that :lol:

    Thanks
 
 
 
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