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Complex Line Integral (should be easy)

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Using a partial fraction decomposition, show that if z lies in the right half plane and C is the line segment from 0 to z, then

    C dz/(z2+1) = i/2 Log(z+i) - i/2 Log(z-i) + π/2​

    2. Relevant equations

    Log(z) = ln(z) + i Arg(z) (maybe relevant?????)

    3. The attempt at a solution

    C dz/(z2+1) = -1/2 ∫C dz/(z+1) + 1/2 ∫C dz/(z-1) = -1/2 Log(z+1) + 1/2 Log(z+1) .......... This isn't looking promising. Suggestions?
     
  2. jcsd
  3. Nov 22, 2011 #2
    You're not doing the partial fractions correctly. You got i's in there you know. Factor z^2+1 and get:

    [tex]\frac{1}{z^2+1}=\frac{1}{(z+i)(z-i)}[/tex]

    now do the partial fractions and then consider the analyticity of the antiderivative along the integration path. Did I just say that? Jesus, what does that even mean? Where are the branch cuts in all of this? branch-points? Do we even need them? Why don't we just try an easier one first like just:

    [tex]\int_0^1 \frac{dz}{z-i}[/tex]

    that 's easy right? It's just:

    [tex]\int_0^1 \frac{dz}{z-i}=\text{mylog}(z-i)\biggr|_0^1[/tex]

    with mylog just being an analytic part of the multivalued function log(z-i). Say we define it as:

    [tex]\text{mylog}(z-i)=\ln|z-i|+i\theta,\quad -\pi<\theta\leq \pi[/tex]

    where theta is the argument (within that range above, of the quantity z-i. So z starts at 0 so then the argument of (0-i) is just -\pi/2. That's within range. Now say z=1/4 +1/4 i so arg of 1/4-3/4 i is what? I don't know exactly. You figure it out but it's in the fourth quadrant so still in range. Same way with the entire length of the integration path: the argument is always within that range above and never crosses the branch-cut along the negative real axis so mylog is analytic along the entire path of integration so therefore I can evaluate the integral by evaluating the antiderivative at it's end points and write:

    [tex]\int_0^1 \frac{dz}{z-i}=\text{mylog}(z-i)\biggr|_0^1=\text{mylog}(1-i)-\text{mylog}(-i)[/tex]

    You not gettin' confussed with that mylog thing right? It's just log ok, or rather the piece of it I define above. You can do the other one. Also, see this thread about using parts of log. Make sure you understand what's going on.

    https://www.physicsforums.com/showthread.php?t=552090
     
    Last edited: Nov 22, 2011
  4. Nov 22, 2011 #3
    I see what's going on now!
     
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