What is the residue at ##z=0## for ##\frac{1}{z^3}+e^{2z}##?

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Homework Help Overview

The discussion revolves around finding the residue of the function ##\frac{1-cos2z}{z^3}## at the point ##z=0##. The subject area involves complex analysis, specifically the evaluation of residues at poles.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the order of the pole and the implications for calculating the residue. There is an exploration of the differentiation process and the correct application of the residue formula.

Discussion Status

Some participants are questioning the initial approach to the residue calculation, particularly regarding the inclusion of the pole's order and the function's structure. There is an acknowledgment of confusion and a shift in focus to a different problem by one participant.

Contextual Notes

Participants are navigating the complexities of residue calculation, including the differentiation of functions and the correct identification of terms in the formula. There is a mention of discrepancies between personal calculations and results obtained from computational tools.

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Homework Statement



Find the residue of ##\frac{1-cos2z}{z^3}## at ##z=0##

Homework Equations



##Res=\frac{1}{n!}\frac{d^n}{dz^n}[f(z)(z-z_0)^{n+1}]## Where the order of the pole is ##n+1##

The Attempt at a Solution



Differentiating ##(1-cos2z)z^3## twice, leaves me with zeros against every term giving me ##0## for a residue which is incorrect. Mathematica gives me 2 for a residue. Since ##z_0=0## I am only left multiplying the top function by z. I have no idea what I am missing here.

What am I doing incorrectly?

Thanks
Chris
 
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What is the order of your pole?
 
3rd order. I think I made a mistake that the 1/z^3 factor was not included in f(x). If I include it, it cancels out the z's outside of the cos2z function altogether. In that case, the part that gets differentiated and evaluated is just 1-cos2x.

Thanks,
Chris
 
kq6up said:
3rd order.
It is not. Don't forget the numerator.
 
I got it. I having trouble with a different problem now. A new thread is in order for that one.

Thanks,
Chris
 

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