What contours will allow the integral to equal 0

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SUMMARY

The discussion focuses on determining the contours, C, that satisfy the Cauchy Integral Theorem for the integral \(\oint \frac{exp(\frac{1}{z^{2}})}{z^{2}+16} dz = 0\). It is established that the contours must avoid the singular points at \(z = \pm 4i\) and \(z = 0\). Participants suggest exploring combinations of contours that exclude these points while still adhering to the theorem's requirements.

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Homework Statement


Determine the contours, C, that will follow Cauchy Integral Theorem so that [itex]\large \oint \frac{exp(\frac{1}{z^{2}})}{z^{2}+16} dz =0[/itex]


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The Attempt at a Solution


The curve can't include z=±4i and z=0 but how do I come up with a curve that does not include those points? Am I just suppose to look for the domain?
 
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dla said:
The curve can't include z=±4i and z=0
You're certainly OK if it does not include any of those, but maybe by including some combination of them you would still get zero. Not sure whether the question expects you to consider that.
 

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