# Integrating Around “Poles”?

In complex analysis and the calculus of "residues," Cauchy's integral theorem gives a "shortcut": The integral is $2\pi i$ times the sum of the "residues." This works because there are "singularities," in the area in question.

Normally, Green's theorem (real case) is a fairly cut and dried matter. But if there is a "singularity" at say, $(0,0)$ then you need to multiply by $2 \pi$ to get the value of the integral.

These two phenomena look suspiciously similar, except for the fact that in the real case, you multiply by $2\pi$, and in the complex case, by $2 \pi i$. Are they, in fact, somehow connected? Or is this a "false" analogy that happens to be coincidental?

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For simplicity suppose that $f$ has a pole located at the origin and let us evaluate the integral $$\oint_{S^1(\epsilon)} f(z) dz$$ over a circle with radius $\epsilon$. This can be parametrized by $z = \epsilon e^{i\phi}$ and one has ${\rm d}z = i z {\rm d} \phi$. This is where the $i$ comes from. From the point of view of the vector analysis ${\rm d} z$ is a tangent (co)vector field (check for yourself that at the point $z = (x,y)$ the tangent field points into the direction given by $iz = (-y,x)$). In other words, $i$ is not that important: we can always pretend that we don't know what complex numbers are and work instead with pairs of real numbers and operations defined on them.
Factors such as $2 \pi$ that these integrals have in common come from the fact that we often integrate over some circles (more generally, spheres). This is natural because we want to enclose the singularity so that it doesn't spoil the calculation. By doing so, we obtain a surface integral over the sphere. When we are finished computing the integral, we can send the radius of the sphere to zero and the resulting integral will be proportional to the area of the sphere times the value of the function (or its gradient, or flow, etc.) over the sphere which will be constant (provided the function is well-behaved).