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How do you determine the "period" of a complex logarithm as a multivalued function in an arbitrary (real or complex) base?

I apologize in advance if my terminology is incorrect, but let me illustrate what I'm talking about. Take $log_e(z)$ for any (non-zero) complex number $z$:

  • $log_e(-1) = \pi i + n2\pi{i}$ for any integer $n$
  • $log_e(2 + 4i) = 1.498 + 1.107 + n2\pi{i}$ for any integer $n$

So $log_e(z)$ is periodic over $2\pi{i}$. I understand how this arises with $log_e$, due to $z = re^{iθ}$, but I don't know how to determine this.

Specifically: how do I figure this out for an arbitrary base $b$ in $log_b(z)$? Is it easier if $b$ is a positive number greater than 1? What about when $b$ is an arbitrary complex number?

I have tried starting with $log_b(z) = log_e(z)/log_e(b)$ and expanding everything out and sort of get an answer by playing around with the results and testing against numeric computations on the computer, but I feel like there has to be a nice, simple, closed form equation to compute this. I'm probably missing something obvious, but I don't know what it is.

EDIT: Revisiting my attempts, what I have so far:

  • Given: $z = re^{iθ}$, $log_e(z) = log_e(r) + iθ + n2\pi{i}$ $\forall n \in Z$
  • Then: $log_b(z) = log_e(z)/log_e(b) = (log_e(r) + iθ)/log_e(b) + (n2\pi{i})/log_e(b)$ $\forall n \in Z$
  • The periodicity would be the last term above $2\pi{i}/log_e(b)$. This didn't seem to always be right ... but see my answer below.
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It's not $\log_b(z)$ that is periodic, it's $b^z$. That is (using $\ln$ for any particular branch of $\log_e$), $b^{z+p} = e^{(z+p)\ln b}$ while $b^z = e^{z \ln b}$, so $b^{z+p} = b^z$ iff $e^{p \ln b} = 1$, i.e. iff $p$ is an integer multiple of $2 \pi i/\ln b$. –  Robert Israel Mar 7 '12 at 21:23
    
@Robert Thanks, I think you've confirmed what I was getting in my attempts, but for some reason my verification attempts where failing. From what you've just said, I think it was because I was assuming that $b^{log_b(z)} = log_b(b^z)$ but that's not really true, is it? –  wjl Mar 7 '12 at 21:33
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1 Answer 1

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With help from Robert's comment, I think I worked with out:

  • Given: $z = re^{iθ}$, $log_e(z) = log_e(r) + iθ + n2\pi{i}$ $\forall n \in Z$
  • Then: $log_b(z) = log_e(z)/log_e(b) = (log_e(r) + iθ)/log_e(b) + (n2\pi{i})/log_e(b)$ $\forall n \in Z$
  • The periodicity would be the last term above $2\pi{i}/log_e(b)$

This didn't seem to always be right, so I thought I had messed it up. However, Robert correctly pointed out in his comment that $log_b(z)$ isn't period -- it's $b^z$ that's periodic. Although, I knew this (although I still don't know what to call a function that is periodically multi-valued in it's range, rather than simply periodic), I made a mistake related to this when trying to verify my results, assuming that $b^{log_b(z)} = log_b(b^z)$ in all cases.

Again, I'm probably not using the right terminology, but the answer to my question appears to be:

$result\_periodicity\_for\_log_b(b) = 2\pi{i}/log_e(b)$ for any complex number base $b$.

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