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Complex integral

  1. Oct 18, 2005 #1
    Hi,

    I'm doing the following as an exercise to try and get my head around complex numbers. Specifically, I need to understand what it means to take the natural log of a complex number and what it involves.

    Say I wanted to integrate 1/ (1 +x^2) dx

    I know this is arcTan(x).


    I can also write the integral as dx / [ (1 + ix)(1 - ix) ], which can be solved with partial fractions to arrive at 0.5 Ln [(1 + ix) / (1 - ix) ].

    Now it should follow that the 2 results are equivalent. How do I got from Ln to arctan? Any hints?

    Mant Thanks for your help,
     
  2. jcsd
  3. Oct 18, 2005 #2
    [tex]\tan{x}=\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}[/tex]
     
  4. Oct 18, 2005 #3
    Looks like you missed a factor of -i of. When you integrate something like [itex]\frac{1}{1+ix}[/itex], you will get something like the result [itex]-iLn(1+ix)[/itex]. Now go and compare the Taylor Series of arctan(x) and your result :wink:
     
  5. Oct 18, 2005 #4
    Gotcha.

    Many thanks.
     
  6. Oct 18, 2005 #5

    George Jones

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    Sorry, LateX preview is not working for me.

    Consider a complex number z = a + ib. In polar form, z = re^{i theta} with r = sqrt{a^2 + b^2} and theta = tan^(-1){b/a}. Now take |z| = r = 1, so b = sqrt{1 - a^2}. Take the ln of both sides of e{^i theta} = a + i sqrt{1 - a^2}.

    i theta = ln(a + i sqrt{1 - a^2})

    i tan^(-1){sqrt{1 - a^2}/a} = ln(a + i sqrt{1 - a^2})

    i tan^(-1){x} = ln{a(1 + ix)}, where x = sqrt{1 - a^2}/a

    Now invert the x equation to find a and you're home.

    Regards,
    George
     
  7. Oct 19, 2005 #6
    Hi George,

    Thanks for the reply but I am still stuck when I use your method. It makes perfect sense but I cant see where I am going wrong. I am doing the integral that I mentioned in my first post; but basically end up with the integral equal to tan^{-1} [ -2x/(1-x^2) ] Sorry, latex doesnt seem to be working for me!

    A brief outline:

    I re-write the integral as

    dx / (1-ix)(1+ix)

    Using partial fractions this comes up to:

    0.5/(1-ix) dx + 0.5/(1+ix) dx

    I do the integral,

    0.5.i. Ln[(1-ix) / (1+ix)]

    then write everything as r.e^{i.theta} . r turns out to be 1 as your mentioned in yoru post, but I end up with something like

    -0.5 arctan{ -2x/(1-x^2) }

    Can you sugegst where I might be going wrong?

    Thanks,
     
  8. Oct 19, 2005 #7

    George Jones

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    Squaring the second equation gives

    a^2 x^2 = 1 - a^2

    a^2 (1 + x^2) = 1

    a^2 (1 +ix)(1 -ix) = 1

    a = [(1 +ix)(1 -ix)](-1/2)

    Using this in the first equation above gives

    i tan^(-1){x} = ln{[(1 +ix)(1 -ix)](-1/2) (1 + ix)}

    i tan^(-1){x} = ln{[(1 + ix)/(1 - ix)]^(1/2)}

    i tan^(-1){x} = 1/2 ln{(1 + ix)/(1 - ix)}

    tan^(-1)(x} = -i/2 ln{(1 + ix)/(1 - ix)}

    tan^(-1)(x} = i/2 ln{(1 - ix)/(1 + ix)}

    This is the same result as yours, but achieved without integration, i.e., the 2 different methods give consistent results.

    Regards,
    George
     
  9. Oct 19, 2005 #8
    much appreciated
     
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