Complex logarithm function

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  1. hopinmad's Avatar
    1 01-08-2012  18:08
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    Complex logarithm function
    Hi,

    The complex (natural) logarithm of z=/=0 is defined as log(z)=log(|z|)+ i arg(z), where log (|z|) is the normal (natural) logarithm for real positive numbers, for some choice of argument for z.

    Let R denote the non-positive real axis. Then, on C-R (by which I mean, the set of complex numbers but not including R), we can choose a choice of argument for each z such that log(z) is continuous (and analytic). For example, choose arg(z) to always lie in [-pi,pi).

    Now, suppose I have a function f(z), defined on some domain D, and I want to consider its complex logarithm. Suppose that, for all z in D, f(z) is never a member of R. What can I say about log(f(z))? I want to say its continuous (and analytic) on D, somehow. But a choice of argument has to come in again. Is it correct to say that, whatever choice of argument I make for f(z), I have a continuous (and analytic) function of z, on the whole of D? I'm dubious because I feel I should be considering the argument of z, not f(z).

    If it's not clear, it's probably because I'm not sure what's going on.

    Now suppose f(z)=exp (g(z)), for some analytic in D function g. I want to say log (exp (g(z)))=g(z), but, again, there are issues with choosing arguments. All I can say with certainty is that log(exp(g(z)))=g(z)+2pi ki, for some integer k. Is it true, somehow, that there is a choice of argument (I'm not sure if I mean a choice for g, for exp (g), or for z here) such that log(exp(g))=g will hold on the whole of D?

    As you can tell, I'm very confused about it. I'd be grateful to anyone casting any light on the matter for me.

    Thanks.
  2. nuodai's Avatar
    2 01-08-2012  18:42
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    Re: Complex logarithm function
    If f is analytic on D \subseteq \mathbb{C} and f(D) \subseteq \mathbb{C} - \mathbb{R} then in particular your choice of \log is analytic on f(D) and hence \log \circ f is analytic on D. Remember, all the logarithm in \log \circ f 'sees' is what f throws into it, and so for the purposes of choosing an appropriate branch cut, you just need to consider the values that f can take.

    It is not true in general that \log \exp w = w (for any given choice of branch cut), but it is true in general that \exp \log w = w, no matter what that choice of branch cut. Now \arg \exp w = \mathfrak{I}w, so setting w=g(z) we have that as long as \mathfrak{I}g(z) is contained in some (open) interval of length 2\pi, you can choose a branch of \log such that \log \exp g(z) = g(z). However, if the image of \mathfrak{I}g ventures outside such an interval, you don't have left-cancellation of \exp g(z) by \log.
    Last edited by nuodai; 01-08-2012 at 18:48.
  3. hopinmad's Avatar
    3 01-08-2012  19:21
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    Re: Complex logarithm function
    Thanks, excellent reply.

    If the image ventured outside such an interval, we get Im(g(z))=/= arg (g(z)), no? Have I understood?
  4. nuodai's Avatar
    4 01-08-2012  19:31
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    Re: Complex logarithm function
    (Original post by hopinmad)
    Thanks, excellent reply.

    If the image ventured outside such an interval, we get Im(g(z))=/= arg (g(z)), no? Have I understood?
    Sorry, I messed up a bit in saying that. What I mean is that \mathfrak{I} w = \arg w + 2n \pi for some n \in \mathbb{Z}, so we can write \mathfrak{I}g(z) = \arg g(z) + 2n(z) \pi, where n(z) \in \mathbb{Z} is an integer-valued function of z. But n(z) is constant if \mathfrak{I}g(z) lies in an appropriate* interval determiend by your choice of argument (say \arg z \in [0, 2\pi)), and so in this case we can choose a suitable branch for the logarithm.

    *By 'appropriate' I mean that if you have \arg z \in I then we need \mathfrak{I}g(z) \in I + 2k \pi for some k \in \mathbb{Z}.
    Last edited by nuodai; 01-08-2012 at 19:35.
  5. hopinmad's Avatar
    5 01-08-2012  19:37
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    Re: Complex logarithm function
    I'm not sure if I've understood your last post. I'll say the following, and if I've misunderstood I'll read your last post again.

    If we (can) choose a domain D'\subseteq D such that we have

    arg[g(z)] \in I,

    where I is some (closed-open) interval of length no greater than 2pi,

    then arg (e^{g(z)}) = Im(g(z)) on D' and hence

    log (exp (g(z))) on D'.

    (Also, how do I get LaTeX into my replies?)


    What can you say about that?
    Last edited by hopinmad; 01-08-2012 at 19:45.
  6. nuodai's Avatar
    6 01-08-2012  20:22
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    Re: Complex logarithm function
    (Original post by hopinmad)
    (Also, how do I get LaTeX into my replies?)
    Press 'Quote' underneath my post to see how I did it.

    (Original post by hopinmad)
    I'm not sure if I've understood your last post. I'll say the following, and if I've misunderstood I'll read your last post again.

    If we (can) choose a domain D'\subseteq D such that we have

    arg[g(z)] \in I,

    where I is some (closed-open) interval of length no greater than 2pi,

    then arg (e^{g(z)}) = Im(g(z)) on D' and hence

    log (exp (g(z))) on D'.
    Sorry, it's a bit involved, so I'll try and spell it out in baby steps :p:

    When we say \arg z what we mean is some \theta \in \mathbb{R} such that z = \left| z \right| e^{i\theta}. There are infinitely many such choices, but they all differ by some integer multiple of 2\pi. If we work with functions we need to choose an interval for the argument to lie in, and this interval has length 2\pi. Say this interval is I \subseteq \mathbb{R}, then for now we'll denote the choice of argument by \arg_I z. So for example, \arg_{[0,2\pi)} (-i) = \dfrac{3\pi}{2} but \arg_{(-\pi,\pi]}(-i) = -\dfrac{\pi}{2}.

    The different choices of interval I correspond to different choices of 'branch' of the logarithm. We'll write \log_I z = \log \left| z \right| + i\arg_I z. In order for us to have analyticity we must take I to be open, which means that there is a half-line starting at 0 in the Argand diagram along which \log_I is not defined. So if you choose I=(0, 2\pi) then all the numbers with argument 0, i.e. the non-negative reals, have no logarithm with this choice of branch. Here our 'branch cut' would be along \mathbb{R}_{\ge 0}, so that the domain of definition for \log_{(0, 2\pi)} is \mathbb{C} - \mathbb{R}_{\ge 0}.

    So far so good (I hope!).

    So say we've picked our interval I = (\theta, \theta+2\pi) for some \theta \in \mathbb{R}. Then \log_I is analytic on \mathbb{C} - \mathbb{R}_{\ge 0}e^{i\theta} =: \mathbb{C}_{\theta}. That is, \log_I analytic on the set \mathbb{C}_{\theta} of complex numbers except for 0 and those with argument \theta.

    Now if D \subseteq \mathbb{C} is some domain and f : D \to \mathbb{C} is an analytic function, in order for \log_I \circ f to be defined and analytic, we must have f(D) \subseteq \mathbb{C}_{\theta}. We don't require D \subseteq \mathbb{C}_\theta, since the input of \log_I in \log_I(f(z)) is whatever f spits out, which is the range of f, and so all that we require is that f(D) \subseteq \mathbb{C}_{\theta} and then we're good.

    Now suppose g : D \to \mathbb{C} is a function and we want \log_I \exp g(z) = g(z) for all z \in D. Well notice that for any w with \arg w \ne \theta we have \arg_I \exp w = \mathfrak{I}w + 2n \pi for that n \in \mathbb{Z} (depending on w) for which \mathfrak{I}w + 2n\pi \in I. Since n depends on w, if we allow w to vary, say w=g(z), then n will depend on z, so that \arg_I \exp g(z) = \mathfrak{I}g(z) + 2n(z)\pi.

    Now, we want a way of keeping n=n(z) constant (or, better, constantly equal to zero). Then by adjusting our choice of I, from I to I + 2n\pi if required, we'd have \arg_I \exp g(z) = \mathfrak{I}g(z). Well the way we keep n constant is by making sure that \mathfrak{I}g(z) stays within a some interval of length 2\pi for all z; if this is the case, then we can take this interval to be our I and we get n=0 in the above, and all's dandy. If \mathfrak{I}g(z) strays outside of such an interval, then the argument will 'wrap around' causing a discontinuity, and this discontinuity will correspond to an increment/decrement in n(z).

    So provided we have the above conditions, so that \arg_I \exp g(z) = g(z), we then have \log_I \exp g(z) = \log_i \left| \exp g(z) \right| + i \arg \exp g(z). Since \left| \exp g(z) \right| = \exp \mathfrak{R}g(z), we see immediately that \log_I \exp g(z) = \mathfrak{R}g(z) + i\mathfrak{I}g(z) = g(z).

    So in order for \log \exp g(z) = g(z), we must restrict g : D \to \mathbb{C} to some subdomain D' \subseteq D for which there exists an open interval I \subseteq \mathbb{R} of length 2\pi with \mathfrak{I}g(z) \in I for all z \in D'. If this holds, then \log_I \exp g(z) = g(z) for all z \in D'.

    Hopefully this wan't too waffly.
    Last edited by nuodai; 01-08-2012 at 22:19.
  7. hopinmad's Avatar
    7 01-08-2012  23:04
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    Re: Complex logarithm function
    Not at all, it was brilliantly explained. I think I was just about there (I should have said, in the post just before, \mathfrak g(z){I} instead of argg(z)) but this post has clarified it 100%. Taking complex logarithms is a sensitive issue.

    Thank you very much.
  8. hopinmad's Avatar
    8 01-08-2012  23:05
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    Re: Complex logarithm function
    the above should say, "Im (g(z) instead of arg (g(z))"
  9. nuodai's Avatar
    9 01-08-2012  23:15
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    Re: Complex logarithm function
    (Original post by hopinmad)
    Taking complex logarithms is a sensitive issue.
    You've hit the nail on the head :p: I wish you luck.
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