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Integration and log of a complex number

  1. Oct 18, 2011 #1
    Hey,

    I know the answer to this integral is 2ipi as it was given but I trying to find out how its 2ipi.

    Here is my working

    [PLAIN]http://img707.imageshack.us/img707/1681/unledrny.jpg [Broken]

    I've been looking at this for ages and I can't work out what i've done wrong


    thanks,
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 18, 2011 #2
    The complex logarithm is ambiguous (as you have already found out) and therefore not always suitable. I would check http://en.wikipedia.org/wiki/Complex_logarithm#Constructing_branches_via_integration very carefully, I'm not sure if you can actually use the logarithm here.
    You can always substitute [itex]u=e^{it}[/itex].
    What will you get as an expression then?
    What will be the integration path?
    Can you spot a singularity? Calculate its residue and you're done :)
     
  4. Oct 18, 2011 #3
    If I change variables to u = e^(it) will i get,

    du/dt = ie^(it), so dt=1/(ie^(it)

    so int (2u/(2u-1))e^(-it) du, this is where im going to stuff up alot, i did it two ways, since u =e^(it) they cancel, so you get in int (2/(2u-1)) du then that integral ends up as the nasty log in it again so then i tried
    leaving the e^(it)
    and i got the nasty log thing + u - 1/2

    I've been looking over my notes and the only stuff about logs he has given use is that log(z)=log|z| + i arg(z), i tried in mathematica and wolframalpha seperately and they gave different answers haha, wolfram gave 2ipi thought
     
  5. Oct 18, 2011 #4
    Forget logarithm! I mean, I can't really tell you why, but maybe someone else can.

    So, let's go through it
    [tex]
    \int_0^{2\pi} \frac{2ie^{it}}{2e^{it}-1} dt
    [/tex]
    [tex] u=e^{it}[/tex]
    [tex] du = i e^{it} dt [/tex]
    So the integral becomes
    [tex] \int_S \frac{2}{2u -1} du [/tex]
    where you integrate over the unit circle.
    Now, the function has a simple pole at u=1/2. Now calculate the residue of this point and you will get that it's 1. Apply the residue theorem and you will get that the value of the integral is 2 pi i.
     
  6. Oct 18, 2011 #5
    Oh right,

    Sorry this is the first time i've every read about residue or the residue theorem.

    This question was actually a question thats really easy to answer with cauchy's integral theorem, but they asked to do it without using it, I know the answer to the question is 6ipi using CIF and i end up with 3[log(ew)] between 0..2Pi

    its just integrating (z^2+z+1)/(z-1) dz over a circle of radius 2, so like 1 step using the cauchy integral formula haha

    What i've done is used long division, found most things cancel and end up with the 3log buisness
     
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