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I'm trying to determine the contour integral of the equation above. I first split the fraction into $g(z)=\frac{z}{z+3}\frac{1}{z+1}$

Then, since $z_0=-1$ is enclosed in the circle $|z+1|=2$, where -1 isn't analytic, I want to focus on $f(z)=\frac{z}{z+3}$.

Next, $\oint \frac{z}{z+3}dz=2\pi if(-1)=2\pi i\frac{-1}{-1+3}=-\pi i$

Are these steps correct? Is there a way to verify this?

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  • $\frac{z}{z+3}\frac{z}{z+1}$ gives $\frac{z^2}{z^2 + 4z + 3}$, which is not the same as $f(z)$. – Arthur Nov 13 '16 at 15:34
  • You are correct -let me change that- should have been 1/(z+1) – Learner Nov 13 '16 at 15:35
  • So there's a simple pole on z₀ = -1, and z₀ = -3. You're correct that -1 is in the circle. However, -3 is on the circle, and should be accounted for. – Kaynex Nov 13 '16 at 15:46
  • @kaynex Then I'd have two residues, -1/2 (as I showed) and 3/2, so the contour integral would be $-\pi i$ and $3\pi i$ – Learner Nov 13 '16 at 15:53
  • I'm new to complex analysis myself, and haven't seen a problem with a pole on the contour before. I'm looking up other questions on the subject. (http://math.stackexchange.com/questions/1075962/pole-on-the-contour-using-the-residu-theorem-what-is-this-formula-of-plemelj) It seems complicated to evaluate a pole on the contour. – Kaynex Nov 13 '16 at 15:58
  • @Kaynex wouldn't the poll that is on the contour, -3 in this case, not be included since we're talking about simple closed path by definition of simple? – Learner Nov 13 '16 at 16:05
  • Remember that residue theorem (or Cauchy's integral formulas, whichever you're working with) depends on deformation of contours, and the ability to put a circular region around every pole in the contour.

    Deformation of contours works by including a new circular region in the original contour, then showing both contours have the same value, since Cauchy-Goursat is satisfied.

    In this case, if we leave the pole on the contour out, then Cauchy-Goursat is not satisfied. So, we have to include it. However, we also can't draw a circular region around it to apply residue theorem.

     – Kaynex Nov 13 '16 at 16:13

1 Answers1

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HINT:Integral $I=2πi×Res[f(z),z=-1]+πi× Res[f(z),z=-3]$

Edit-A more general form of Cauchy's residue theorem:

$\int_C f(z)dz=2πi.\sum Res[f(z);z=z_k]+πi.\sum Res[f(z);z=z_c]$

where $z_k$ is a pole inside contour $C$ and $z_c$ is a pole on the boundary of contour $C$.

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Nitin Uniyal
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