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# Complex logarithm function Tweet

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1. 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. Re: Complex logarithm function
If is analytic on and then in particular your choice of is analytic on and hence is analytic on . Remember, all the logarithm in 'sees' is what throws into it, and so for the purposes of choosing an appropriate branch cut, you just need to consider the values that can take.

It is not true in general that (for any given choice of branch cut), but it is true in general that , no matter what that choice of branch cut. Now , so setting we have that as long as is contained in some (open) interval of length , you can choose a branch of such that . However, if the image of ventures outside such an interval, you don't have left-cancellation of by .
Last edited by nuodai; 01-08-2012 at 18:48.
3. Re: Complex logarithm function

If the image ventured outside such an interval, we get Im(g(z))=/= arg (g(z)), no? Have I understood?
4. Re: Complex logarithm function

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 for some , so we can write , where is an integer-valued function of . But is constant if lies in an appropriate* interval determiend by your choice of argument (say ), and so in this case we can choose a suitable branch for the logarithm.

*By 'appropriate' I mean that if you have then we need for some .
Last edited by nuodai; 01-08-2012 at 19:35.
5. 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. Re: Complex logarithm function
(Also, how do I get LaTeX into my replies?)
Press 'Quote' underneath my post to see how I did it.

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

When we say what we mean is some such that . There are infinitely many such choices, but they all differ by some integer multiple of . If we work with functions we need to choose an interval for the argument to lie in, and this interval has length . Say this interval is , then for now we'll denote the choice of argument by . So for example, but .

The different choices of interval correspond to different choices of 'branch' of the logarithm. We'll write . In order for us to have analyticity we must take to be open, which means that there is a half-line starting at 0 in the Argand diagram along which is not defined. So if you choose 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 , so that the domain of definition for is .

So far so good (I hope!).

So say we've picked our interval for some . Then is analytic on . That is, analytic on the set of complex numbers except for 0 and those with argument .

Now if is some domain and is an analytic function, in order for to be defined and analytic, we must have . We don't require , since the input of in is whatever spits out, which is the range of , and so all that we require is that and then we're good.

Now suppose is a function and we want for all . Well notice that for any with we have for that (depending on ) for which . Since depends on , if we allow to vary, say , then will depend on , so that .

Now, we want a way of keeping constant (or, better, constantly equal to zero). Then by adjusting our choice of , from to if required, we'd have . Well the way we keep constant is by making sure that stays within a some interval of length for all ; if this is the case, then we can take this interval to be our and we get in the above, and all's dandy. If 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 .

So provided we have the above conditions, so that , we then have . Since , we see immediately that .

So in order for , we must restrict to some subdomain for which there exists an open interval of length with for all . If this holds, then for all .

Hopefully this wan't too waffly.
Last edited by nuodai; 01-08-2012 at 22:19.
7. 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, instead of arg) but this post has clarified it 100%. Taking complex logarithms is a sensitive issue.

Thank you very much.
8. Re: Complex logarithm function
the above should say, "Im (g(z) instead of arg (g(z))"
9. Re: Complex logarithm function
Taking complex logarithms is a sensitive issue.
You've hit the nail on the head I wish you luck.