In complex analysis we study a term RESIDUE of a function given by some formulas. While going through its meaning I found that it means left out term or remainder kind of thing. so I was wondering why is this term given such a terminology, can anyone enlighten me towards connection of this mathematical term with its literal meaning.
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It's what's left after integrating over a closed path enclosing the singularity. – Daniel Fischer Sep 02 '13 at 13:57
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@Daniel can u plz elaborate how it gives the leftout portion after integrating – user Sep 02 '13 at 14:01
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Not sure what you mean with "left out". It's what is "left over", what remains. Everything else vanishes when integrating. – Daniel Fischer Sep 02 '13 at 14:05
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See also http://math.stackexchange.com/questions/255612/intuitive-explanation-of-residue-theorem-in-complex-analysis. – lhf Oct 24 '16 at 00:45
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Assume that $f$ is holomorphic on $\mathbb{D}\backslash\{0\}$ and has a pole at $0$. Then $f$ has a Laurent expansion $f(z)=\sum_{k\geq n}a_kz^k$, where $n\in\mathbb{Z}$. One can see, using for example the Cauchy integral formula, that $\int_\gamma z^kdz=0$ for any closed simple curve around the origin and for any $k\neq -1$. The problem is then at $-1$. It can be shown that $\int_\gamma\frac{1}{z}dz=2\pi i$ for a closed simple curve around $0$, and in general for any curve $\frac{1}{2\pi i}\int_\gamma\frac{1}{z}dz$ is the winding number of $\gamma$. Therefore, if $\gamma$ is any closed simple curve around $0$, we have that $$\frac{1}{2\pi i}\int_\gamma f(z)dz=\frac{1}{2\pi i}\sum_{k\geq n}a_k\int_\gamma z^kdz=a_{-1}.$$ This is the residue. Therefore, $a_{-1}$ is what's "left over" after integrating.
rfauffar
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Of course this works on any simple connected domain with a point removed, I just used the disk for simplicity. – rfauffar Sep 02 '13 at 14:12