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Complex Problem

How does someone know a priori (I mean without doing calculations, as here!) that $A,B,C$ are the residues?

I mean of course if we use the standard formula for residues on the RHS we get $$\lim_{z\to 1}(z-1)\left(\frac{A}{z-1}+\frac{B}{z-w}+\frac{C}{z-\bar{w}}\right) = A$$ and you can do the same for the others and check they are indeed the residues.

Is this though a general "formula"/"recipe"? whenever we have a fraction where the denominator can be factorised in terms of exponent one, can we automatically say that then the residues are the coefficients of the partial fractions?

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  • $\begingroup$ Remember that one way of finding residues is to consider the Laurent series expansion of the function in question about the considered points. Here, the points are the three 3rd roots of unity, and is is obvious that you can write the integrand fucntion in the form above. Hence, the integrand is already(!) written in its Laurent series expansion about any of the (simple) poles, with $a_{-1} = A, B, C$, respectively (where $a_{-1}$ denotes the first coefficient of the Laurent series expansion about each pole, hence the residue!). $\endgroup$ – ComplexFlo Dec 28 '16 at 14:10
  • $\begingroup$ @FBiersack I thought about that! However I don't understand the following: it seems indeed written in its Laurent Series expansion, however normally you find it around one point only. So normally you'd find the laurant series around, say $1$ and write it up, involving only powers of $(z-1)$. However here we have al three together, how is this acceptable? As a counterexample take the Taylor series of $e^x = \sum_{n=o}^{\infty}\frac{x^n}{n!}$ around zero, so involves only $(x-0)$ terms, whereas around $x = 1$ it will involve $(x-1)$ terms only, right? $\endgroup$ – Euler_Salter Dec 28 '16 at 14:14
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If $a$ is a pole of $f$, then the residue at that pole is the coefficient of $(z-a)^{-1}$ in the Laurent expansion of $f$ centered at $z=a$. One often obtains the Laurent expansion by "tricky" rewriting of the function, as in this case where you used partial fraction decomposition.

The other terms are holomorphic there since they represent poles at other locations, so they don't contribute to the residue. That's why you can deal with one term at a time.

Example: The function $f(z)= \frac7{z-1} + z^2+3$ has residue $7$ at the pole $z=1$. We didn't write out the whole Laurent series explicitly, but we could: $$f(z) = \frac7{z-1} + z^2+3$$ $$= \frac7{z-1} +\underbrace{((z-1)^2 +2z -1)}_{z^2} + 3$$ $$= \frac7{z-1} + (z-1)^2 +2z +2$$ $$=\frac7{z-1} + (z-1)^2 + \underbrace{(2(z-1) + 2)}_{2z} + 2$$ $$=\frac7{z-1} + (z-1)^2 + 2(z-1) + 4$$ Now everything is in terms of powers of $z-1$, but the holomorphic part doesn't affect the residue, so why bother?

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  • $\begingroup$ oh wait, so what you are saying is: when I'm looking at $z = 1$ , say, the other two fractions are not part of the series? So if I consider the series around $z=1$, I'll consider only the first term as Laurent Series (LS), if I consider the expansion around $z=w$ I'll consider only the second term as LS and same for the third? $\endgroup$ – Euler_Salter Dec 28 '16 at 14:16
  • $\begingroup$ All the three terms together, do they still form a Laurent Series? Or does it become a Laurent Series only when I choose a point ($z=1$, $z=w$ and so on) to consider the series around? $\endgroup$ – Euler_Salter Dec 28 '16 at 14:17
  • $\begingroup$ No, they don't form a Laurent series. But you don't really need the whole series, just that particular term, to get the residue at that pole. The other terms do indeed contribute to the series, but they will only contribute to the holomorphic part (nonnegative powers of $z-a$) in the Laurent series. I will add a simple example to the answer to avoid extended comment. $\endgroup$ – MPW Dec 28 '16 at 14:40

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