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The residue theorem that states that if

a) $U$ is a simply connected and open subset of the complex plane,

b) $a_1,\dots,a_n$ are finitely many points of $U$,

c) and $f$ is a function which is defined and holomorphic on $U\backslash \{a_1,\dots,a_n \}$,

d) $\gamma$ is a rectifiable and positively oriented curve in $U$ which does not meet any of the $a_k$, and whose start point equals its end point,

then $$ \oint_\gamma f(z)\, dz = 2\pi i \sum_{k=1}^n \operatorname{Res}( f, a_k ). $$ Question: What is the intuitive explanation of this theorem ?

I appreciate a explanation like geometry ( whit Java Applet ? ). But algebric explanation are welcome. I know the proof of this theorem, I'm just trying to understand the intuition ( if exists ) this theorem.

I'm motivated to get an intuition of this theorem because he gave a wonderful explanation of the intuitive of an part of fundamental theorem of calculus using the properties of the telescopic sums, mean value theorem and Riemann sum whit partition $\{a=x_0<x_1<\dots<x_n=b \}$ of interval: \begin{align} F(b)-F(a)= & \sum_{k=1}^{n}[F(x_k)-F(x_{k-1})] & \mbox{ telescopic sum } \\ = & \sum_{k=1}^{n}F^\prime(x_k^*)[x_k-x_{k-1}] & \mbox{Mean Value Theorem} \\ = & \sum_{k=1}^{n}F^\prime(x_k^*)\Delta x_k & \mbox{Riemann sum} \\ \approx & \int_{a}^{b}F^\prime(x)\, dx & \end{align}

It makes me think that there may be an explanation as elegant as this (but not necessarily following the same reasoning) in understanding the theorem of residues.

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See also math.stackexchange.com/questions/4054/…. –  lhf Dec 11 '12 at 0:07

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up vote 1 down vote accepted

The algebraic intuitive explanation is that for all $n\in \mathbb Z$, the functions $z^n$ integrate to $0$ around a simple closed loop containing $z=0$, except when $n=-1$, when the integral is $2\pi i$.

From this observation it follows that if $f$ is a meromorphic function with one simple singularity at $z=a$ then $f(z)=g(z)/(z-a)$ where $g$ is holomorphic and when you expand $f$ in a Laurent series around $z=a$ (by using the Taylor series of $g$) you get one term $g(a)/(z-a)$ plus other terms that do not contain $z-a$ to a negative power, that is, a power series around $z=a$. When you integrate $f$ in that form, all terms but one become zero. The one that is left (hence the name residue) is the one corresponding to $g(a)/(z-a)$, whose integral is $2\pi i g(a)$. This is the Residue Theorem. It needs to be extended to handle higher-order singularities and multiple singularities but the essential idea is the one above.

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You may also enjoy Needham's Visual Complex Analysis. –  lhf Dec 11 '12 at 0:06

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