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Partial Fractions with Complex Numbers

  1. May 12, 2013 #1
    Let's start with:

    $$ \int \frac{dx}{1+x^2} = \arctan x + C $$

    This is achieved with a basic trig substitution. However, what if one were to perform the following partial fraction decomposition:

    $$ \int \frac{dx}{1+x^2} = \int \frac{dx}{(x+i)(x-i)} = \int \left[ \frac{i/2}{x+i} - \frac{i/2}{x-i} \right] \, dx $$

    I believe that this is justified because plugging in any real number ##a## into the integrand would simplify to a real number (namely ##1+a^2##).


    $$ \int \frac{dx}{1+x^2} = (i/2) \ln |x+i| - (i/2) \ln |x-i| + C $$

    In conclusion:

    $$ \arctan z = \frac{i}{2} \ln \left| \frac{z+i}{z-i} \right| $$

    The right side should yield real value outputs for real value inputs.

  2. jcsd
  3. May 12, 2013 #2


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    Actually once you go to complex numbers you don't want the absolute values in your logarithms anymore. Have you studied complex analysis at all (it makes the explanation of what's going on easier)?
  4. May 12, 2013 #3
    Not at all. I am, however, obsessed with complex numbers, and am taking an introductory level complex analysis course next year.

    Is it correct though? Also, reflecting now, there should be a " ## + 2\pi i k ## on either side of the equation, right?
  5. May 12, 2013 #4


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    I don't understand why you would want that. (and if it's on both sides, why you can't cancel it)

    The basic thing is that you can define logarithm to be nice and diferentiable for all complex numbers except for a single half-axis. That half-axis is usually picked to be the negative real numbers. You can define logarithm differently to get it to be defined everywhere except for a different half-axis, for example the positive real numbers.

    The usual definition of logarithm that is defined for positive numbers extends to something which is defined everywhere except for negative real numbers, with no problem. And for this definition of logarithm,
    [tex] \frac{d}{dz} \log(z) = \frac{1}{z} [/tex]
    So the antiderivatives of [itex] \frac{1}{z} [/itex] are just log(z)+C in this whole region.

    When you take your complex analysis class you will probably learn about this - defining a function everywhere except for a line is what's known as picking a branch cut for the function.

    Have you seen the expression/understand the expression [tex] e^{i\theta} = \cos(\theta) + i \sin(\theta)[/tex]
    Because if you do, then explaining why log needs a branch cut becomes fairly simple
  6. May 12, 2013 #5
    Well, when I said "either", I meant "one". The natural log is the inverse to e^z, which has period ##2\pi i##. If ##u = (i+z)/(i-z)## (from the equation above), then:

    $$ \ln u = y $$

    $$ u = e^y $$

    $$ u = e^y \cdot e^{2 \pi i k}, \quad k\in \mathbb{Z} $$

    $$ \ln u = y + 2 \pi i k $$

    $$ y = \ln u + 2 \pi i k $$

    Also, what's the problem with defining the natural log for negative reals? Can't you just do some algebra:

    $$ \ln (-x), \quad x>0 $$

    $$ = \ln(-1) + \ln x $$

    $$ e^{i \pi + 2 i \pi k} = e^{i \pi (1+2k)} = -1 $$

    $$ \implies \ln(-1) = i \pi (1+2k), \quad k \in \mathbb{Z} $$

    Thus, is it safe to say that:

    $$ \ln (-x), \quad x>0 = \ln x + i \pi (1+ 2k), \quad k \in \mathbb{Z} $$
  7. May 12, 2013 #6


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    Combining your first and your last lines you get
    [tex] y = y+2\pi i k [/tex]
    Which is clearly false. For each choice of k you can define a natural log function, but once you pick a k you have to stick with it. This is because ln is what's called a multi-valued function. If you want to do analysis, you have to pick a branch (this is related to the branch cut I mentioned earlier) which means pick a specific choice of values and never look back. For ln this means pick a choice of k (usually k=0) and work with that

    The problem with ln for negative reals is that I want ln to be a differentiable function. That means I pick a choice of k (to make it a function) and then I try to make it differentiable. This means that I can't include all the numbers on the unit circle because
    [tex] \ln(e^{i\theta}) = i\theta [/tex]
    causes a lot of problems if I start with [itex] \theta = 0[/itex] and go around the unit circle. It's continuous, but when I get back to the number 1 I have [itex]\ln(1) = 2\pi i[/itex] which is problematic - the function cannot be defined continuously if I allow myself to go around the whole circle. So you have to pick some axis to declare a branch cut - where you aren't allowed to evaluate the function at, and you just remove from the domain - and the most common choice is the negative real axis. This means that if we throw away the negative real axis, we can define ln(z) so that it is a differentiable function everywhere else
  8. May 12, 2013 #7
    Ok, that makes sense.
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