Finding residue of a function(complex analysis)

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SUMMARY

The discussion centers on finding the residue of the function \(\frac{e^{iz}}{(1+9z^{2})^{2}}\) using the residue theorem in complex analysis. The user encountered difficulties with the Laurent series expansion, resulting in a zero in the denominator. The correct approach involves recognizing the pole of order 2 at \(z = i/3\) and applying the residue formula: \(\text{Res}(f,i/3) = \lim_{z \to i/3}\frac{d}{dz}\left( (z-i/3)^2 \frac{e^{iz}}{(1+9z^2)^2} \right)\).

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Homework Statement


I need to find the residue of \frac{e^{iz}}{(1+9z^{2})^{2}} so I can use it as part of the residue theorem for a problem.


Homework Equations


Laurent Series
R(z_{0}) = \frac{g(z_{0})}{h^{'}(z_{0})}


The Attempt at a Solution


I tried using the laurent series but after expanding I got a 0 in the denominator.
For the second equation I used I also got a 0 in the denominator and I don't believe it converges. Any help would be appreciated. Thanks!
 
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Look at the inverse of the function (call it f(z)), you have a zero of order 2. So you have a pole of order 2 for your function. The formula for this goes like ## \text{Res}(f,i/3) = \lim_{z \to a}\frac{d}{dz}\left( (z-i/3)^2 \frac{\exp (i z)}{(1+9z^2)^2} \right) ##

you can do the others

## \text{Res}(f,a) = \lim_{z \to a} \frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}} (z-a)^m f(z) ##

If the pole is of order m. I think. Maybe look that up.
 

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