Pole of a function, as a geometric series

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SUMMARY

The function f(z) = 1/(1 + e^z) has poles determined by the condition when the denominator equals zero, specifically at z = ln(-1). The discussion highlights the expansion of the function as a geometric series, represented as 1 - e^z + e^(2z) + ..., which does not exhibit poles in its series form. The key takeaway is that the poles occur at z = iπ, where the function diverges, and the series expansion should be centered around this point for further analysis.

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  • Complex analysis fundamentals
  • Understanding of poles and singularities
  • Knowledge of geometric series expansions
  • Familiarity with the exponential function e^z
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cragar
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Homework Statement


Determine the order of the poles for the given function.
f(z)=\frac{1}{1+e^z}

Homework Equations

The Attempt at a Solution


I know if you set the denominator equal to zero
you get z=ln(-1)
But if you expand the function as a geometric series ,
1-e^{z}+e^{2z}...
I don't see how there is a pole in the geometric series expansion , there is no division by zero. [/B]
 
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You want to expand as a series consisting of powers of ##z-z_0## about the point ##z_0 = i\pi##.
 

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