Calculating the Residue of ##\frac{1}{(x^4+1)^2}## at Double Poles

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SUMMARY

The residue of the function ##\frac{1}{(x^4+1)^2}## at its double poles, located at ##z=e^{\frac{i \pi}{4}}##, ##z=e^{\frac{3i \pi}{4}}##, ##z=e^{\frac{5i \pi}{4}}##, and ##z=e^{\frac{7i \pi}{4}}##, can be calculated using the limit formula for higher-order poles in complex analysis. The correct residue value is ##\frac{3}{16 \sqrt{2}}+\frac{3}{16 \sqrt{2}}i##. While the standard residue formula is not applicable due to the nature of the double pole, the limit formula simplifies the process. The discussion highlights the algebraic complexity involved in deriving the residue through Laurent series expansion.

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Homework Statement


How would I calculate the residue of the function

##\frac{1}{(x^4+1)^2}##

Homework Equations

The Attempt at a Solution


So I have found that the poles are at

##z=e^{\frac{i \pi}{4}}##
##z=e^{\frac{3i \pi}{4}}##
##z=e^{\frac{5i \pi}{4}}##
##z=e^{\frac{7i \pi}{4}}##

I tried calculating this by finding its laurent series around each of the poles, but it was very algebraically heavy and I could not get the correct answer of

## \frac{3}{16 \sqrt{2}}+\frac{3}{16 \sqrt{2}}i##

I feel there must be an easier way. The standard residue formula also does not work here (i'm assuming because its a double pole)

Any help would be extremely appreciated!
 
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