Ian H
Lv 7

Complex integral around unit circle at origin for 1/z is 2𝛑i, For 1/z^n it is zero as per Cauchy s first even though pole there. Why?

The function f(z) = 1/z, is

not analytic at the origin; it has a type of singularity there called a pole.

Accepting that result and the reason given, it seemed reasonable that

if f(z) = 1/z^n, around a unit circle, then ∮(1/z^n)dz would also NOT be zero.

But the result of calculation is that ∮(1/z^n)dz = 0, (unless n = 1)

4 Answers

  • 1 year ago
    Favourite answer

    Cauchy's result isn't nonzero because there's a pole there, it's nonzero because the function (in some sense, described below) has a 1/z term in it, and that's the only term that integrates not to zero.

    Let's do specific examples, first we'll integrate something like dz / z^2 which is an example of your general case for n = 2 to show that it has to be zero. Then, we'll show why dz / z isn't zero. It will then be clear why this is the only integral of this form that's not zero.

    A path integral literally means we count up contributions of the function evaluated at points in this path only, this path is a circle which you can parameterize as e^{it} for t going from 0 to 2pi, and the function is 1 / z^2 (for this example). This means we want to evaluate the integral when z = e^{it} over this range. Let's do the integral

    integral dz / z^2 over unit circle

    which, as mentioned, is just an expression asking us to count up all values only over the unit circle; this is z = e^{it}, hence dz = i e^{it} dt, so we have

    = integral i e^{it} dt / e^{2 i t} over (t = 0 to t = 2pi )

    = integral i * e^{-it} dt over (t = 0 to t = 2pi)

    = e^{-it} evaluated over (t = 0 to t = 2pi)

    = 0

    Why is this zero? The integral ended up being something where the function in the integrand has the same starting point value as its ending point value. It vanishes just as we are used to in real-calculus. What's so special about the n = 1 case?

    integral dz / z over unit circle

    = integral i e^{it} dt / e^{it} over (t = 0 to t = 2pi )

    = integral i * dt over (t = 0 to t = 2pi)

    = i * t over (t = 0 to t = 2pi)

    = 2 pi i

    Why isn't that zero? Because the exponentials cancelled this time. That's the only time kind of integral would ever be nonzero, if the exponentials cancelled. So, for n =/=1 (not just for n = 2), we expect some exponential to hang around and this integral therefore vanishes.

    The n = 1 case is a very special case, it's related to something a "residue" because (as it sounds) it's the only "part" of a function that "survives" a closed-loop integration in the complex plane.

    What do I mean by "part" of a function and "survives"? Well, you have some function, like f(z) = e^{z} / z^5, and you want to integrate it over any closed path, you can put in the path z = whatever and try to do the integration just as we did above (before we had a simple path, unit circle has z = e^{it}). Even if the path is simple, you usually can't do that integral as it stands (you can't do this one either), so how do you proceed? You can consider how else you might attempt it, maybe a different "view" this integration would be obvious: enter the famous power series representation of functions. You are probably used to Taylor/MacLaurin series, these are like sin(x) = x - x^3 / 3! + x^5 / 5!. You could integrate sin(x) to get -cos(x) directly, there's no difficult in doing this directly in this case, but it's nice to also notice that (if needed) you did have the option of integrating the power series form instead term-by-term (cf. Fubini's theorem for validity of interchanging operations if curious).

    So, go back to the complex-valued case, we don't know how to integrate e^z / z^5 over some closed contour in the complex plane the straightforward way (sticking in z = whatever and integrating). How about you get a power series form of this? If you do such a thing, you could integrate this term-by-term just the same, and that'd be the end of it. It really is that easy. This is a little different (work it out if you're curious) because the function is not as nice as sin(x) was (what's not "nice" about it is that it has a point z = 0 where it blows up). Because of this, and this happens for any function that's like this, the power series you get will end up having negative and positive powers**, not just positive powers like the series for sin(x) did. So, you'd get something like

    f(z) = a5 / z^5 + a4 / z^4 + ... + a1 / z + b0 + b1 * z + b2 z^2 + ...

    where the a's and b's are just constants you'd arrive at by writing down the power series for e^z and multiplying through by 1 / z^5. We are now tasked with integrating this term-by-term. Each term looks like your prototype 1 / z^n. So, what's the answer for integrating this over a closed loop?

    integral f(z) dz = integral [a5 / z^5 + a4 / z^4 + ... + a1 / z + b0 + b1 * z + b2 z^2 + ...] dz

    ........................= 0 + 0 + ... + 2 * pi * i * a1 + 0 + 0 + 0 + ...

    ........................= 2 * pi * i * a1

    the term a1 is called a "residue" and what I meant by it's the only thing that survives an integration (note that the b0 term vanishes because in integrating this we can't forget about the dz which has some form like i e^{it} and that integrates to zero over a closed loop). So, the 1 / z term is acknowledged to be very special. It forms the basis of so-called residue theorem. Anyway, I've talked too much.

    ** this is called a Laurent series

  • Vaman
    Lv 7
    1 year ago

    It is a tricky question. You have to integrate

    dz/z^n. Write z=r e^(i a). I am using a in place of theta.

    dz = r i da e^(ia) da e^ i(-na)/r^n

    integral will be r^(-(n-1))e^(-i(n-1)a)) da. Integrate it over 0 to 2 pi. You get

    r^(1-n)(1/(-i(n-1)) (exp(-2pi(n-1)-1)

    You see that irrespective r values, the integral vanishes if n is not 1.

  • JOHN
    Lv 7
    1 year ago


    Attachment image
  • Anonymous
    1 year ago

    school hasn’t started yet kira

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