- Residues as "climbing between the values of a multivalued antiderivative"
- Winding numbers and the residue theorem
- All residues are logarithmic residues: the Laurent series
- "Proving Laurent series": Laurent series as Fourier series
- How the residue theorem gives life to $\pi$
The story so far: if a function has no screw-up points within a closed contour, its integral on that contour is zero. But if it does, it may not be.
And by a screw-up point, we just mean a point at which the function isn't (or rather cannot be -- this is what we call a non-removable singularity) holomorphic.
But why? Let's look at the antiderivative of $1/z$ -- $\mathrm{log}(z)$. It's fundamentally a multivalued function -- and here's what's interesting: a loop around the origin isn't actually a loop on this graph -- it brings you to a higher level than you were previously (specifically $2\pi i$ higher than you were previously).
And you can kinda see why this comes about -- the derivative of this function not being holomorphic at 0 is encapsulated by the fact that the function is all torn up at 0 -- its slope must be different in different directions, because you have multiple planes stuck to that point. This is the idea behind a branch point -- a point such that the function is discontinuous when going about an arbitrarily small circuit about the point.
One can now start to see how integrals that do loop around the origin behave -- for starters, the winding number of the contour corresponds to the number of levels you climb during the integral. Also, the distance between levels should depend purely on the local nature of the function around the branch point (because the antiderivative has a defined derivative (the function $f$, in this case $1/z$), so the spacing must remain constant). By similar reasoning, encircling multiple poles means climbing all their levels (so the total distance climbed adds up).
This, above, is the residue theorem. For a function $f$, a simply connected open set $U$ such that $f$ is holomorphic on $U-\{a_1,\dots a_k\}$, and a closed contour $\gamma$ contained in this punctured set:
$$\oint_\gamma f(z) dz = \sum_{k=1}^n \mathrm{W}(\gamma, a_k)\mathrm{Res}(f,a_k)$$
Where $\mathrm{Res}(f,a)$ is the residue of $f$ at $a$, which is the "local quantity" that equals the "distance between layers" of the multivalued antiderivative.
This is, obviously, awesome. But thinking about generally handling and computing these residues $\mathrm{Res}(f,a)=\oint_{\circ}f(z)dz$ leads us to wonder about the general nature of branch cuts.
Here's an idea: we know exactly what the residue for $f(z)=z^{-1}$ at 0 is -- it's $2\pi i$. One also easily sees that the residue of $f(z)=z^n$ for any other $n$ is zero (the antiderivative $z^{n+1}/(n+1)$ does not have branch cuts).
I wonder if -- like how Taylor series allow us to represent a function near a general point $a$ as a sum of $(z-a)^n$s for nonnegative $n$ -- we could represent a function near a singularity $a$ as a sum of $(z-a)^n$s for all integer $n$s.
I wonder if all "serious" singularities are just ultimately $1/z^k$-style singularities.
This is, in fact, what is known as the Laurent series -- a representation $f(z)=\sum_{n=-\infty}^\infty c_n (z-a)^n $. Then the distance climbed by $f(z)$ is equal to the distance climbed by the $(z-a)^{-1}$ term, which is just $2\pi i c_{-1}$.
So the point of this "Laurent series interpretation" of the residue is this:
- The key conceptual takeaway from the Laurent series is that all branch cuts of the antiderivatives of holomorphic functions are "logarithmic". (This does not, e.g. apply to the branch cut of $\sqrt{z}$, as its derivative isn't holomorphic.)
- When actually calculating residues, we can find the Laurent series by other means (such as by patching together different Taylor series) and use its $c_{-1}$ coefficient as the residue.
But to actually "prove" this interpretation means to:
- Write down what its coefficients should be -- this is our candidate series which justifies using it to calculate residues.
- Show that this candidate is a valid Laurent series, i.e. that it actually converges to $f(z)$ on some region.
- Show that it is the valid Laurent series, i.e. that the Laurent series is unique. So we can calculate the Laurent series however we want and use its coefficients to calculate residues.
The Laurent series is not the Taylor series, and the nonnegative coefficients of the Laurent series are not generally the coefficients of the Taylor series. So its coefficients cannot be interpreted as higher derivatives of the function or anything (that doesn't even make sense at that point).
The first one is easy. Obviously we need ${c_{ - 1}} = \frac{1}{{2\pi i}}\oint_\circ {f(z)\,dz}$. Analogously by considering this expression for the function $\frac{{f(z)}}{{{{(z - a)}^{n + 1}}}}$ to extract the other coefficients, we see that:
$${c_n} = \frac{1}{{2\pi i}}\oint_\circ {\frac{{f(z)}}{{{{(z - a)}^{n + 1}}}}\,dz} $$
The second and third are actually challenging -- but there's a remarkable fortunate observation one can make. The fact that the Laurent series is defined on an annulus (each of the power series has a radius of convergence, which puts a maximum bound on both $z$ and $|z|$) is incredibly suggestive of describing a periodic function. Indeed, a variable substitution $z=\rho e^{i\theta}$ turns the Laurent series into a Fourier series. The properties 2 and 3 then follow from properties of the Fourier series.
This "Fourier series" interpretation of the Laurent series also makes it easy to see Cauchy's integral formula and holomorphicity implies analyticity.
By the way, I would argue that the notion of a residue and the Laurent series is the fundamental source of the importance of $\pi$. Perhaps the standard motivation for $\pi$ is that $2\pi i$ is the period of the exponential function. This is equivalent to saying it's the residue of the logarithm function. The existence of Laurent series expansions -- i.e. all branch cuts being logarithmic in nature -- is what gives importance to these residues.
Exercise: prove the Cauchy differentiation formula -- for a holomorphic function $f$,
$$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}dz $$
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