Contour Integrals and Residues

Question

I'm trying to figure out what it is all about, but my mind is blowing up. First of all, I have turned back and looked at the general definitions of integrals. Then I have looked to line integrals. Basically, the integral is helping us to find the area between two points along the curve. For example when force $F$ is constant but height of the object varies point to point, we can use the integral to find the total work done by $F$.

The line integral is helping us to find the total work done by any vector or scalar fields along any path. Arch length formula gives us the length of curve. So if all these definitions are Ok, please help me to understand about Contour and Residues? Why do we need to use them? Is there any physical meaning or are we just expanding our number set to use The Euler formula and obtain the result easily?

Answer 1

An analytic function $f:\ \Omega\to{\mathbb C}$ can be viewed as a kind of "complex force field" in $\Omega$, so that it makes sense to integrate $f$ along curves $\gamma\subset\Omega$.

A physical force field $F$ is conservative (meaning: it conserves energy) when the integral of $F$ along any closed curve $\gamma\subset\Omega$ is zero. Now it is a basic property of analytic functions $f$ that they are locally conservative, meaning that integrals along "short" closed curves are zero. The reason for this is the following: The "complex curl" $${\partial f\over\partial\bar z}$$ of such a function is identically zero – this is the content of the CR equations.

But "in the large" such a complex force field $f$ need not be conservative, as is exemplified by the function $$f(z):={e^z\over z^3}$$ in the punctured complex plane. When $\partial D$ is the counterclockwise oriented unit circle then $$\int_{\partial D} f(z)\ dz=\pi i\ne0\ .$$ In this example the non-conservativeness of $f$ can be precisely localized: It is the pole, an isolated singularity, of $f$ at $0$. But that's not all: The value of the integral need not be computed the hard way; it can be extracted from the Laurent expansion of $f$ at $0$, which is given by $$f(z)={1\over z^3}+{1\over z^2}+{1\over 2z}+{1\over 6}+{z\over 24}+\ldots\quad.$$ The residue of $f$ at $0$ is on the one hand the coefficient ${1\over2}$ of the ${1\over z}$-term in this expansion, and on the other hand it is the value of the above integral, divided by $2\pi i$. That these two numbers coincide is the content of the residue theorem (which, of course, covers much more complicated cases).

Answer 2

In complex analysis, you might want to think of contour integrals as anti-derivatives. This helps to explain why the path can be moved (as long as it does not pass over a singularity) without altering the integral; it is just the difference of the anti-derivative at the endpoints.

Another way to express this is using Green's Theorem applied to a complex differentiable function: $$ \oint_{\partial\Omega} f(z)\,\mathrm{d}z =\iint_\Omega(if_x-f_y)\,\mathrm{d}x\,\mathrm{d}y\tag{1} $$ The Cauchy-Riemann Equations are simply $$ if_x=f_y\tag{2} $$ Applying $(2)$ to $(1)$ yields Cauchy's Integral Theorem: $$ \oint_{\partial\Omega} f(z)\,\mathrm{d}z=0\tag{3} $$ when $f$ is holomorphic (i.e. satisfies the Cauchy-Riemann Equations) in $\Omega$.

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Thus, the difference of the integral along the green path and the integral along the red path is the integral along the blue path. According to $(3)$, that is $0$.

There are points at which otherwise holomorphic functions go bad. These are singularities. The residue is $\frac1{2\pi i}$ times the integral counter-clockwise around such a point.

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Note that the difference between the integrals along the green and red paths is the integral along the blue path, taking the integral in along the bottom of the blue line and out along the top (the integrals along the lines cancel, being in opposite directions). The blue contour is the boundary of the "C"-shaped area which avoids the singularity, so the integral along the blue contour is $0$. Therefore, the integral along the green path equals the integral along the red path. Thus, it does not matter which contour around the singularity we take, the residue will be the same.