I've heard of residues in complex analysis, contour integration, etc. but all I really know it to be is the $c_{-1}$ term in the Laurent series for a function. Is there some sort of intuition on what a "residue" actually is? The terminology makes it seem like something left over, or something like that.

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    $\begingroup$ It is what is left when you integrate around a point. $\endgroup$ Commented Aug 18, 2015 at 3:30
  • $\begingroup$ Or it's what's left over after you cancel out everything you can by subtracting off the derivative of some other Laurent series. $\endgroup$
    – user14972
    Commented Aug 18, 2015 at 3:42
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    $\begingroup$ It sounds like you are after some intuition. The intuition here comes from the Residue theorem: $\int f(z) dz = 2\pi ic_1$ for a simple closed curve containing only a lone singularity. The holomorphic portion of $f$ adds 0 to the integral by Cauchy's theorem. The other terms in the laurent expansion also provide 0 because they are the derivatives of single-valued functions. The only term that provides any value - the residue of integration about the curve - is the $1/z$ term. $\endgroup$ Commented Aug 18, 2015 at 3:51
  • $\begingroup$ @PaulSinclair could you explain to me why the other terms in the laurent expansion provide zero because they are the derivatives of single-valued functions? first time learning complex analysis and this is not clear to me. Thank you $\endgroup$
    – Monolite
    Commented Apr 10, 2017 at 15:12
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    $\begingroup$ @Monolite - because you are integrating about a closed curve. By the FTC (applied though the definition of integration along a curve), the integral will be the difference of the antiderivative at the two endpoints. But if those endpoints are the same, the difference is $0$. However, the natural logarithm is not single-valued. While the curve goes around the singularity, it picks up $2\pi i$ instead of returning to its original value. $\endgroup$ Commented Apr 10, 2017 at 17:13

3 Answers 3


The holomorphic functions have an extraordinary property: if you compute an integral along a path, the value of the integral does not depend on the path !

More precisely, if the function is holomorphic everywhere inside a closed path, the integral is just zero. But if the function has poles (zeroes at the denominator, $c_{-k}$ terms in the Laurent series), every pole brings a nonzero contribution called its residue. You can shrink the path as much as you want, even turning it to infinitesimal circles around every pole, provided you keep the poles in.

So the residues are what is left (as regards integration) after you removed all the holomorphic parts of the domain.

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  • $\begingroup$ Surely I would not vote in favour of this beautiful answer that I am particularly interesting. $\endgroup$
    – Sebastiano
    Commented Aug 17, 2020 at 19:14

A function $f$ analytic in a full disk $D_r$ around $a$ can be written as derivative of some other function $F$ in $D$:$$f(z)=F'(z)\quad(z\in D_r)\ .$$ If $f$ has an isolated singularity at $a$ one may still ask whether $f$ has a primitive $F$ in the punctured disk $\dot D_r:=D_r\setminus\{a\}$, in other words: whether the ODE $y'=f(z)$ has a solution in $\dot D_r$. It turns out that the sole obstruction to the solvability of this problem is the residue $${\rm res\,}_a(f):={1\over2\pi i}\int_{\partial D_\rho} f(z)\>dz,\qquad\rho<r\ .$$ If this residue is $\ne0$ no solution exists.

In my history of math book (by Moritz Kline) I read that the name of "residue" has been introduced by Cauchy in his Exercices de mathématique (1826–30) in connection with an integral similar to the above, but around a rectangle.

  • $\begingroup$ Very deep. But is this really the original reason for naming the residue so? $\endgroup$ Commented Aug 18, 2015 at 10:53

This doesn't exactly answer your question, but a related question is "what's so special about the $z^{-1}$ term in the Laurent expansion?" The answer can be stated very simply and can be understood without needing complex numbers: Every other term in the Laurent series integrates to a power function. That one doesn't.


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