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Old October 27th, 2009, 02:10 PM
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Default complex integration

Integrate the funtion around 4 different contours
g(z)=z^2+1/z^2-1
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  #2  
Old October 27th, 2009, 02:49 PM
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Originally Posted by bookie88 View Post
Integrate the funtion around 4 different contours
g(z)=z^2+1/z^2-1

Just like that?? Well, then integrate about 4 contours none of which contains in its interior or on its boundary neither the point z = 1 or z = -1 and then your function if analytic and thus the integrals all equal zero...simple!

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Old October 28th, 2009, 04:26 AM
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Originally Posted by bookie88 View Post
Integrate the funtion around 4 different contours
g(z)=z^2+1/z^2-1
the significance of different contours lies in whether or not your contour encircles (if its a closed contour) the singular points of g(z)

singular points are just where g(z) is undefined, so here, where the denomenator equals 0
where z=1 or -1

so a contour that doesnt encircle these points, would be
\gamma = 3+e^{i\theta}

integrating g(z) around this contour gives 0

this is by the residue theorem, closed contour integration theorem... something along those lines

for a contour that encircles the singularities
e.g.
\gamma = 2e^{i\theta}
you have to calculate the residues of g(z) at the points of singularities
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Old October 28th, 2009, 05:36 AM
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Originally Posted by walleye View Post
the significance of different contours lies in whether or not your contour encircles (if its a closed contour) the singular points of g(z)

singular points are just where g(z) is undefined, so here, where the denomenator equals 0
where z=1 or -1

so a contour that doesnt encircle these points, would be
\gamma = 3+e^{i\theta}

integrating g(z) around this contour gives 0

this is by the residue theorem, closed contour integration theorem... something along those lines

for a contour that encircles the singularities
e.g.
\gamma = 2e^{i\theta}
you have to calculate the residues of g(z) at the points of singularities
We all understand what you're saying. However, the question does not specify which four contours to use so the question is answered by choosing any four contours at all, including the four suggested in post #2 .... (A flaw in either the original question or in how the question has been posted ....)
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