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I'm wondering if there's a way to guess the value of complex contour integral of some meromorphic function or at least its sign, anything useful really. I've been thinking about exploiting vector field intepretation. Consider a complex function $f(z)=f(x+iy)=u(x,y)+iv(x,y)$ holomorphic on some open subset $D\in\mathbb{C}.$ Notice that vector field $\overrightarrow{F}=(u,-v)^T$ is conservative. It's due to the fact that holomorphic functions suffice Cauchy-Riemann equations, i.e. $$\frac{\partial}{\partial x}u(x,y)=\frac{\partial}{\partial y}v(x,y),$$ $$\frac{\partial}{\partial y}u(x,y)=-\frac{\partial}{\partial x}v(x,y).$$ In 2 dimensions we have $$\mathrm{curl}\overrightarrow{F}=-\left(\frac{\partial}{\partial y}u(x,y)+\frac{\partial}{\partial x}v(x,y)\right)\overrightarrow{k}=0,$$ $$\mathrm{div}\overrightarrow{F}=\frac{\partial}{\partial x}u(x,y)-\frac{\partial}{\partial y}v(x,y)=0,$$ where $\overrightarrow{k}=(0,0,1)^T.$ Means that $\overrightarrow{F}$ is both irrotational and solenoidal. The above is particulary useful when trying to prove Cauchy integral theorem and path-independence of the line integral for holomorphic functions. Taking it further, let us consider some meromorphic function $g(z)$ with one isolated singularity at some $z_0$ then for aribtrary $r>0$ the integral over $\partial S=\{z\in\mathbb{C}:|z-z_0|=r\}$ counter-clockwise is \begin{align} \int_{\partial S}g(z)\mathrm{d}z&=\int_{\partial S}\left(\varphi(x,y)+i\psi(x,y)\right)\mathrm{d}(x+iy)\\&=i\int_{\partial S}\psi(x,y)\mathrm{d}x+\varphi(x,y)\mathrm{d}y+\int_{\partial S}\varphi(x,y)\mathrm{d}x-\psi(x,y)\mathrm{d}y\\&=i\iint_{S}\left(\frac{\partial \varphi}{\partial x}-\frac{\partial \psi}{\partial y}\right)\mathrm{d}x\mathrm{d}y-\iint_{S}\left(\frac{\partial \psi}{\partial x}+\frac{\partial \varphi}{\partial y}\right)\mathrm{d}x\mathrm{d}y\\&=i\iint_S\mathrm{div}\overrightarrow{G}\mathrm{d}x\mathrm{d}y+\iint_S\mathrm{curl}\overrightarrow{G}\circ \overrightarrow{k}\mathrm{d}x\mathrm{d}y, \end{align} where $\overrightarrow{G}=(\varphi,-\psi)^T$. As an example we can look at $g(z)=1/z$, we know the value $\int_{\partial S}g(z)\mathrm{d}z=2\pi i.$ Is there any actual correspondence to the formula I've written? The only I see is that rotation part must be equal to zero. Then all the contribution comes from divergence at $z_0$. Of course curl and divergence can be undefined or infinite at $z_0$, but don't bother for a while and think in sense of limits maybe.

In other words, I'm interested in comprehending and somehow describing perturbation of such vector field caused by the presence of singularity

I suppose there's some mind-blowing interpretation with vortices, sources, sinks and molecules travelling through the field analogous to those appearing in multivariable calculus. I really hope someone happens to see it and is so kind to share his or hers views.

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  • $\begingroup$ The theorem that lets you move from path-integrals to area integrals requires that $\psi$ and $\varphi$ are continuously differentiable over $S$ (or somewhat weaker, depending on the version). This is not the case when $f$ has a singularity. $\endgroup$ Jun 26 '17 at 23:05
  • $\begingroup$ That's right. Nevertheless it still makes sense when thinking in terms of limits. For example take the sequence $g_n=\frac{\overline{z}}{|z|^2+1/n}$ it converges to $g = 1/z$ pointwise and each $|g_n|$ is bounded. Then for each $n$ functions $\varphi_n$ and $\psi_n$ are well-defined on the whole $\mathbb{C}$ hence also line intergrals and vector field exist. For each $g_n$ Green's theorem applies, too. $\endgroup$
    – piwox
    Jun 30 '17 at 14:11
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I've come up with my own idea, the explanation may not be rigorous, but I'm quite convinced it can be made such. My question was restricted to meromorphic functions only, which is in fact misleading. It should be extended to a broader non-holomorphic class.

Consider function $f(z)=\frac{1}{z}$ and sequence of functions $$f_n(z)=\frac{\overline{z}}{|z|^2+\frac{1}{n}}, \quad\quad n\in\mathbb{N}\,.$$ Obviously $f_n\rightarrow f$ pointwise. Each $f_n$ can be naturally decomposed into its real and imaginary part $f_n(x+iy)=u_n(x,y)+iv_n(x,y)$. In a sense it's a way to avoid the presence of singularity, yet managing to capture the desired behaviour. Note that $|f_n|$ is bounded for each $n$ on the whole $\mathbb{C}$. It follows that we can define a vector field $\overrightarrow{F_n}=(u_n,-v_n)^T$ on any subset of $\mathbb{C}$ for every $n\in\mathbb{N}$. By the argument from posted question it's now valid to say $$\int_{\partial S}f_n(z)\mathrm{d}z=i\iint_S\mathrm{div}\overrightarrow{F_n}\,\mathrm{d}x\mathrm{d}y+\iint_S\mathrm{curl}\overrightarrow{F_n}\circ\overrightarrow{k}\mathrm{d}x\mathrm{d}y\,,$$ where $S\subset\mathbb{C}$. Computing appropriate derivatives we arrive with $$\frac{\partial}{\partial x}u_n(x,y)=\frac{y^2-x^2+\frac{1}{n}}{\left( x^2 + y^2+\frac{1}{n}\right)^2},\quad \frac{\partial}{\partial y}u_n(x,y)=-\frac{2xy}{\left( x^2 + y^2+\frac{1}{n}\right)^2}$$ $$\frac{\partial}{\partial x}v_n(x,y)=-\frac{2xy}{\left( x^2 + y^2+\frac{1}{n}\right)^2}, \quad\frac{\partial}{\partial y}v_n(x,y)=\frac{y^2-x^2-\frac{1}{n}}{\left( x^2 + y^2+\frac{1}{n}\right)^2}\,.$$ Hence $$\mathrm{div}\overrightarrow{F_n}=\frac{\frac{2}{n}}{\left( x^2 + y^2+\frac{1}{n}\right)^2},\quad \mathrm{curl}\overrightarrow{F_n}=0\,.$$ Let now focus on contour integral over circle of radius $R$ $$\int_{|z|=R}f_n(z)\mathrm{d}z=\frac{2i}{n}\iint_{|z|\leq R}\frac{1}{\left( x^2 + y^2+\frac{1}{n}\right)^2}\mathrm{d}x\mathrm{d}y=\frac{2i}{n}\int\limits_{0}^{R}\!\!\int\limits_{0}^{2\pi}\frac{r}{\left(r^2+\frac{1}{n}\right)^2}\mathrm{d}\vartheta\mathrm{d}r=$$ $$=\frac{2\pi i}{n}\int_{0}^{{R}^2}\frac{1}{\left(t+\frac{1}{n}\right)^2}\mathrm{d}t=\frac{2\pi i}{n}\left(n-\frac{1}{R^2+\frac{1}{n}} \right)=\frac{2\pi i}{n}\left( \frac{nR^2+1-1}{R^2+\frac{1}{n}}\right)=$$ $$=2\pi i\frac{R^2}{R^2+\frac{1}{n}}\xrightarrow{n\to \infty}2\pi i = \int_{\partial S}f(z)\mathrm{d}z,\quad \forall R>0$$ Of course one will immediately notice that in general a limit of sequence of integrals of functions does not equal an integral of limit of function sequence. Nevertheless it turned out to be true in this specific case and continues to be true for any other fixed example. That's the part where I seem not to be able to state it rigorously, but I suppose that some convergence of measures would be the way.

The intuition standing behind this observation is quite remarkable.

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Here's a picture output with WolframAlpha for $\overrightarrow{F_{100}}$. As expected, there's no vortex and the field diverges evenly around $z_0=0$. In a limiting sense divergence at all the non-zero points vanishes with $n$ growing large and concentrates as a infinite peak at zero - just like the so called $\mathrm{\textit{Dirac's delta}}.$ Therefore my guess is that Dirac's measure would help to prove it formally. Other elegant intuitive way to explain the phenomenon is to imagine there's a source of evenly intensive light at $z_0$. No matter the distance and shape if we surround it by some closed curve, all the density is captured and thus must sum up to the same value $2\pi$. Other example would be straight wall which must capture exactly $i\pi$ of light density - same as half of a circle. This density by some trigonometry is $\frac{i}{1+t^2}$ and its antiderivative then equals $i\mathrm{arctan}(t)$, which gives $\pm i\pi$ as desired when evaluated over any straight line not passing through $z_0$ (plus or minus sign depends on whether singularity is on the left-hand side or on the right-hand side of the line).

Next step shall be driven by the question: what would happen if we'd consider $f(z)=\frac{i}{z}$. The integral by linearity is obviously equal to $-2\pi$, but take a look at another picture.

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Can you guess what happens with the field for $-\frac{i}{z}$ and $-\frac{1}{z}$? Probably it's now obvious. It twists clockwise and converges to zero respectively.

enter image description here enter image description here

That corresponds to the fact that multiplacation of complex numbers of modulus $1$ is rotation. In terms of vectors it curls the field. Contour integral can be interpreted as work of the field on a travelling particle anti-clockwise around circle. Parallel work adds up as the real component and perpendicular as the imaginary one. Then you can easily guess the looks of any 'multiplied' field, it just combines forces. Modulus of the multiplier stands for its power.

The $2\pi$ constant is due to the circle circumference. However, what would happen if we've taken an ellipse to begin with? Nothing much really because of the transition to polar coordinates later on, which would convert it back to a circle anyway. The same would occur with any other closed curve, but integration would just become unnecessarily complicated. The choice of a circle is most natural. What's more that explains why contour integral value of a meromorphic function does not depend on the shape of region of integration

Seeking further correspondence to residue theorem let's check what happens for $\frac{1}{z^2}$ and $\frac{1}{z^3}$ and so on.

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Here the complex plane is divided into even number of cancelling out subfields implying that the integral is equal to $0$, which is in fact true when computing according either to the definition or to given formula.

The above deals with all the negative powers of $z$. Positive ones and zero-th are holomorphic functions, so we know they get integrated to zero also. Notice now that vector fields are additive. Also, any complex function has non-zero radii annulus in which its Laurent series converges. To recall, Laurent series is defined as a power series including negative powers $$f(z)=\sum_{n=-\infty}^{\infty}a_nz^n, \quad a_i\in\mathbb{C}, \quad r<z<R\,.$$

We arrive with a conclusion that contour integral of any Laurent-expandable complex function depends solely on its $-1$st term and whether singularity is contained in the region of integration or not. If there's more than one singularity in such particular region, we know all the divergence and rotation is concentrated at these points so the fields don't interfere with each other and can be simply added with the restriction, that singularities have to be isolated ones, otherwise the limiting argument fails. That's quite exactly what residue theorem states.

The residue itself can be thought of as a setting of the field of $\frac{1}{z}$. It determines the power, sign and curl/divergence ratio. That fact can be used to transform the field to a irrotational normalised one just with multiplying by a single complex constant. Here the interpretation with light source comes in handy.

The formula holds also for any function with continuously differentiable real and imaginary part both bounded on a subset of $\mathbb{C}$ (to suffice Green's theorem assumptions and for existence of vector field).

Well, maybe someone happens to read it through and finds it interesting.

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