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I Contour integration - reversing orientation

  1. Oct 5, 2016 #1
    I have been reading through "Complex Analysis for Mathematics & Engineering" by J. Matthews and R.Howell, and I'm a bit confused about the way in which they have parametrised the opposite orientation of a contour ##\mathcal{C}##.

    Using their notation, consider a contour ##\mathcal{C}## with parametrisation $$\mathcal{C}:\;z_{1}(t)=x(t)+iy(t)\,,\;\;\; a\leq t\leq b$$ Then the opposite curve ##-\mathcal{C}## traces out the same set of points but in the reverse order, and it has the parametrisation $$-\mathcal{C}:\;z_{2}(t)=x(-t)+iy(-t)\,,\;\;\; -a\leq t\leq -b$$ They then go on to say that since ##z_{2}(t)=z_{1}(-t)##, it is easy to see that ##-\mathcal{C}## is merely ##\mathcal{C}## traversed in the opposite sense.

    However, I don't quite see why the contour ##-\mathcal{C}## is parametrised this way? Naively, I would've though it would be something like $$-\mathcal{C}:\;z_{2}(t)=x(t)+iy(t)\,,\;\;\; b\leq t\leq a$$

    Furthermore, how does one show that if two contours ##\mathcal{C}_{1}## and ##\mathcal{C}_{2}## are placed end to end, such that the terminal point of ##\mathcal{C}_{1}## coincides with the initial point ##\mathcal{C}_{2}##, then the integral over the contour ##\mathcal{C}=\mathcal{C}_{1}+\mathcal{C}_{2}## has the property $$\int_{\mathcal{C}}f(z)dz=\int_{\mathcal{C}_{1}+\mathcal{C}_{2}}f(z)dz= \int_{\mathcal{C}_{1}}f(z)dz+\int_{\mathcal{C}_{2}}f(z)dz$$ Would one use the following parametrisations: $$\mathcal{C}_{1}:\;z(t)=x(t)+iy(t)\,,\;\;\; a\leq t\leq b\\ \mathcal{C}_{2}:\;z(t)=x(t)+iy(t)\,,\;\;\; b\leq t\leq c$$ Then, $$\int_{\mathcal{C}_{1}}f(z)dz+\int_{\mathcal{C}_{2}}f(z)dz=\int_{a}^{b}f(z(t))z'(t)dt+\int_{b}^{c}f(z(t))z'(t)dt=\int_{a}^{c}f(z(t))z'(t)dt=\int_{\mathcal{C}_{1}+\mathcal{C}_{2}}f(z)dz=\int_{\mathcal{C}}f(z)dz$$ where $$\mathcal{C}:\;z(t)=x(t)+iy(t)\,,\;\;\; a\leq t\leq c$$
     
  2. jcsd
  3. Oct 5, 2016 #2

    Erland

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    You must have ##-b\le t\le -a##, since ##-a>-b##.
    You can't have ##b\le t\le a##, since ##b>a##.
    Yes, this is the simplest way to do this.
     
  4. Oct 6, 2016 #3
    Apologies, I must've misquoted the book on this bit.
    Is the point that if we want to traverse the contour in the opposite direction then we must choose the parameter such that it ranges in the opposite direction, but since ##b>a## we must consider ##-a## and ##-b## such that ##-b<-a## and ##-b<t<-a##. Since the contour is the same, but simply being traversed in the opposite direction we require the value of ##z_{2}(t)## for ##t=-b## to coincide with the value of ##z_{1}(t)## for ##t=b##, and similarly for all other points in the range, hence we require that ##z_{2}(t)=x(-t)+iy(-t)=z_{1}(-t)##. As such, we have that ##z_{2}(-b)=z_{1}(b)## and ##z_{2}(-a)=z_{1}(a)## as required.
     
  5. Oct 6, 2016 #4

    Erland

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    Yes, that is correct.

    There are other alternatives to parametrize the opposite contour though. One possibility is to choose the parameter interval as the same as the originial one: ##a\le t\le b##, and define ##z_2(t)=z_1(a+b-t)##, but your parametrization is simpler, in my opinion.
     
  6. Oct 6, 2016 #5
    Would be very beneficial to you if you did some experimenting Frank. Sounds like the book meant a closed contour with for example ##x(t)=\cos(t)## and ##y(t)=\sin(t)## but you should confirm that experimentally for example in Mathematica: actually set up a ParametricPlot with those parameters preferably in a Manipulate context and then run the controls to visually convince yourself that the path is reversed. This I believe will help you greatly later on when the question is not so easy and you will have by then experience investigating Complex Analysis experimentally.
     
  7. Oct 6, 2016 #6
    Ah ok. I think I get it now. Thanks for your help!

    Good idea. I'll have a go.
     
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