# Physical Interpretation of the Residue

In my complex analysis course (taught by the physics department no less) we've obviously paid close attention to the residue for solving our problems, but there has been no attempt to try and attach any kind of concrete way to think of the residue, a physical description of the residue in a system. Is there a way of thinking of this, beyond just being a mathematical tool? And similarly, what would be the physical equivalent of our contour integrations?

My intuition is telling me that we could think of the residue as some kind of "behavior source" that buried in some kind of phase space that projects it's behavior into the real axis. Our contour integration is then basically just trying calculate the flux of this dominant behavior over our interval of interest (the projection of the real phenomena in phase space). Even as I say this though, it still feels very vague since I don't have a physical model I can compare this against. Several other answers have spoken of the electrostatic potential but I don't recall any ever really describing what the residue means with respect to our (real-valued) function of interest.

• In what context are you talking about the residue here? Honestly, it is a mathematical tool that is used for a variety of different things in physics, and its physical meaning (if any) can be different depending on what it's being used for. – David Z Feb 8 '15 at 8:47
• I'd like any kind of example that allows for an effective (and accurate)physical description of a residue. – Skyler Feb 8 '15 at 8:49
• I hope the Q&A below shed some light on the residue: physics.stackexchange.com/questions/29532/… – user40602 Feb 8 '15 at 13:01
• I think this Math.SE will help you out. – DanielSank Sep 11 '16 at 23:13

the residue is very similar to div(electric field) , it is like a 1-d counterpart. We know that $\nabla.(r^n)\hat{r}=nr^{n-1}$ at r=0 is 0 for all r$\neq$-2. This is because the $r^{-2}$ 'compensates' the surface $\alpha r^{2}$ and integral is $4\pi$
Similarly when integrating $(z-z_0)^n$, only the $r^-{1}$ term is nonzero and is $2\pi$