Finding Residue of Complex Function at Infinity

AI Thread Summary
The discussion focuses on finding the residue of the function f(z) = z^3 * exp(1/z) / (1 + z) at infinity. The user attempts to express the function in a Laurent series and identifies that the residue corresponds to the coefficient a_{-1}. They expand the series for both the numerator and denominator, leading to a multiplication of the two series to isolate the desired coefficient. The resulting terms are analyzed, yielding a series that includes terms like 1/4! - 1/5! + 1/6! and ultimately suggests a final expression of e^{-1} - 1/2 + 1/3!. The user expresses uncertainty about potential mistakes in their calculations.
MartinKitty
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Hello everyone,
I have a problem with finding a residue of a function:
f(z)={\frac{z^3*exp(1/z)}{(1+z)}} in infinity.
I tried to present it in Laurent series:
\frac{z^3}{1+z} sum_{n=0}^\infty\frac{1}{n!z^n}

I know that residue will be equal to coefficient a_{-1}, but i don't know how to find it.
 
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Expand \frac{z^3}{1+z}=z^3-z^4+z^5-z^6.... The multiply the two series together to find the coefficient you want (as an infinite series).
 
mathman said:
Expand \frac{z^3}{1+z}=z^3-z^4+z^5-z^6.... The multiply the two series together to find the coefficient you want (as an infinite series).
Then i get:
\frac{z^3}{1+z}=(-1)^n*z^{n+3}

and when i multiply I always get {z^3} with some fraction
 
## \left(z^{3}-z^{4}+z^{5}-\cdots \right)\left(1+\frac{1}{z}+\frac{1}{2z^{2}}+\frac{1}{6z^{3}}+\frac{1}{24z^{4}}+\cdots \right)##, the only interested terms are of the forms ##\frac{a_{-1}}{z}##, that are ##\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\cdots ## so it is ## e^{-1}-\frac{1}{2}+\frac{1}{3!} ## (if I did not make mistakes ...)
 
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