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Integration of complex functions

  1. May 28, 2008 #1
    I'm doing a bit of studying on calculus of complex functions.
    The book I'm reading from is "Mathematical Methods For Physicists", and in the course of reading the chapter on 'functions of a complex variable' I have run across the equation

    [tex]\int^{z_{2}}_{z_{1}}[/tex]f(z) dz =[tex]\int^{x_{2},y_{2}}_{x_{1},y{2}}[/tex][u(x,y) + iv(x,y)][dx + idy]

    So to really understand this I tried proving it to myself, that it's true.
    When you substitute u(x,y) + iv(x,y) for f(x) you do get what's up there. so simplify that and you get

    = [tex]\int^{x_{2},y_{2}}_{x_{1},y{2}}[/tex][u(x,y)dx-v(x,y)]+i[tex]\int^{x_{2},y_{2}}_{x_{1},y{2}}[/tex][v(x,y)dy + u(x,y)dy)]

    which is great, but when I try to integrate f(z) = z I get two different answers.
    directly integrating
    [tex]\int[/tex] z dz =
    substitute in x + iy for z
    and you get
    [tex]\frac{x^{2}-y^{2}}{2}[/tex] + ixy

    integrating using the substituted formula
    we get
    [tex]\frac{x^{2}-y^{2}}{2}[/tex] + i2xy

    I can't for the life of me prove to myself that those two equations (the substituted integral and the direct integral) are equal.

    Any help would be greatly appreciated. Thanks
  2. jcsd
  3. May 29, 2008 #2
    Draw a picture of the path that you are integrating. I bet you are going from (0,0) -> (x,0) -> (x,y) then you should see that on the first leg, the imaginary part of the integral is 0, and on the second leg the imaginary part is xy. I think that you didn't make the connection that the y-coordinate on the first leg was 0, you thought it was y and that's why you had the 2?
  4. May 29, 2008 #3
    We were taught, without name of theorem or proof unfortunately :/, that you can let y=0 and x=z and it kicks out the complex representation for your function. What it essentially is saying is that the way an analytic function behaves in the complex domain is the same as it behaves in the real domain.

    I've used it and it works (you get the right answer) but I would love to see why.
  5. May 29, 2008 #4
    You've lost me. I understand how the path is (0,0)->(x,0)->(x,y), but I don't understand how that would leave me with out a two.
    When I posted it was late, and I was sufficiently frustrated with the the latex codes, so neglected to show how I got those two answers, I'll post that now, and perhaps it will make my questions more clear.

    f(z) = z
    z = x + iy
    dz = dx + idy ( <-- for some reason this doesn't look right, this time around, but I can't place it)
    splitting f(x) into real and imaginary parts we get
    f(z) = u(x,y) + iv(x,y)
    u(x,y) = x
    v(x,y) = y

    [tex]\int[/tex]f(z) dz= [tex]\int[/tex]z dz
    =[tex]\int[/tex][u(x,y) + iv(x,y)][dx + idy]
    =[tex]\int[/tex]u(x,y)dx - [tex]\int[/tex]v(x,y)dy +i[[tex]\int[/tex]u(x,y)dy + [tex]\int[/tex]v(x,y)dx]

    then integrating z dz
    [tex]\int[/tex] z dz
    = [tex]\frac{z^{2}}{2}[/tex]
    plug in z = x + iy
    =[tex]\frac{(x + iy)^{2}}{2}[/tex]
    =[tex]\frac{x^{2} - y^{2} + i2xy}{2}[/tex]
    =[tex]\frac{x^{2} - y^{2}}{2} + ixy[/tex]

    but if we integrate with the substituted formula

    [tex]\int[/tex]x dx - [tex]\int[/tex] y dy + i[[tex]\int[/tex] x dy + [tex]\int[/tex] y dx]
    = [tex]\frac{x^{2}}{2} - frac{y^{2}}{2} + i[xy + yx][/tex]
    = [tex]\frac{x^{2} - y^{2}}{2} + i2xy [/tex]
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