# Does the specific contour matter when integrating over a closed loop

I've heard different people say yes and no, so I want to ask it here on math stack, but often, as in with cauchys integral theorem on wiki for instance, the contour is specified as a circle in the complex plane. Could it however be any shape (that's a closed contour) in the complex plane, and if so why?

To me, it doesn't seem to matter, and here's my reasoning: define the arbitrary contour as a parametric function of time, and since it's loop (not sure if there's a more formal definition, but here just take it to mean a closed contour) after some given time (let's call it T) it winds up back were it started at $$t=0$$. The derivative of the curve with respect to time is another complex valued function of time, and since after T the tangent is the same, the derivative is the same at T, meaning the derivative of the curve with respect to time also forms a loop (given of course that the contour is differentiable). So, the opposite also applies. Integrating a loop yields another loop, and integrating the loop over the time it takes to make one revolution (ie a definite integral from 0 to T) is 0 since you haven't moved anywhere since it's a loop. Then, if $$dz=z'dt$$ integrating a complex function over a complex variable is the same as a complex function (since the $$z'$$ that comes out is multiplied and complex numbers are closed under multiplication) integratied over time (a real variable), which is zero unless they cancel. Assuming they don't, symbolically (and I is the antiderivative of Z):

$$\oint_z f(z)dz=\int^T_0 f(z)z'dt=\int^T_0 Zdt=I(T)-I(0)=0$$

$$\frac{1}{z}$$ is an example of what I mean by cancel, if you were to integrate it like this. Making a change of coordinates to polar reveals that regardless of $$r(t)$$ (the loop) it evaluates to $$2\pi i$$.

Again, even if this yields the right results I'm not sure if the intuition's correct. Do people just just use circles as the contour because they're convenient? Thanks in advance, anything that may be helpful would be great!

• Why does putting a loop "in the loop" make the integral invalid? Dec 16, 2018 at 11:49

I am not sure that I understood your question, but it seems to me that you are claiming that if you integrate $$\frac1z$$ along any loop around $$0$$, then you will always get $$2\pi i$$. This is not true. If $$\gamma\colon[0,2\pi]\longrightarrow\mathbb C$$ is defined by $$\gamma(t)=e^{-it}$$, then$$\int_\gamma\frac1z\,\mathrm dz=-2\pi i.$$ So, in that sense at least, the choice of the loop matters.

On the other hand, if we assume that we are not talking about loops on general, but only about simple loops, that is, loops such their winding number with respect to $$0$$ is $$1$$, then, yes, you are right: the shape doesn't matter. It can be a square, a triangle, an ellipse, or whatever you want. Then $$\int_\gamma\frac1z\,\mathrm dz$$ will always be $$2\pi i$$.

• Oh yeah, I forgot about integrating in another direction. But aside from that... does the shape of the path matter? What about any function, not just $\frac{1}{z}$ Nov 22, 2018 at 15:12
• What do you mean by “shape”? Are you thinking about the possibility that the image of $\gamma$ is, say, an ellipse or a square? Nov 22, 2018 at 15:15
• yes, ie the shape of the contour if you graph it. By a different shape that isn't a circle I mean a contour that doesn't look like $e^{it}$. Nov 22, 2018 at 15:40
• I have added another paragraph to my answer. What do you think now? Nov 22, 2018 at 15:45
• And I guess it follows that this goes for any general f(z) without a negative one term in the Laurent series integrating to zero? Nov 22, 2018 at 16:24

You can deform the curve any way you wish, as long as the curve doesn't cross a singularity during its motion.

• Jose Carlos Santos said something about winding number. Why does it matter? Dec 23, 2018 at 20:55
• If it has winding two would it be 4$\pi i$ instead? Dec 23, 2018 at 20:56
• What I said stands. If you cross a singularity, the winding number changes. Dec 24, 2018 at 16:45