Assume $r \in (0,1)$, I want to prove that

$$ \text{Im}\left[r\log\left(\frac{-1 + r}{r}\right)\right]=\pi r$$

This is probably something trivial, but I don't have much experience with complex numbers. Are there any known identities for similar expressions?

As a side note, this logarithm cancels out a specific imaginary part of a Lerch Transcendent in another problem, so I can only guess that it's equal to $\pi i$.

  • 2
    $\begingroup$ What is the meaning $r$ in the both side ? Also $r=0$ is not possible $\endgroup$ Feb 7, 2021 at 8:03
  • $\begingroup$ unless I am missing something, I do not think what you have written is true. Notice if you take e to the power of both sides you clearly get something different. $\endgroup$
    – Moosh
    Feb 7, 2021 at 8:06
  • $\begingroup$ My bad, I will edit the question $\endgroup$
    – runr
    Feb 7, 2021 at 8:08
  • 2
    $\begingroup$ I think it should be $\;r\in(0,1)\;$ , as in the extreme points of the unit interval the expression isn't defined $\endgroup$
    – DonAntonio
    Feb 7, 2021 at 8:16
  • 1
    $\begingroup$ @DonAntonio you're right, thanks a lot! $\endgroup$
    – runr
    Feb 7, 2021 at 8:17

1 Answer 1


It seems to be that by $\;log\;$ you meant the complex log, something which isn't given...and thus:


Now, assuming the usual main branch of the complex logarithm, $\;-\pi<\arg\le\pi\;$ , we get that

$$0<r< 1\implies\frac{-1+r}r<0\implies \arg\frac{-1+r}r=\pi\implies\text{Im}\left(\log\frac{-1+r}r\right)=\pi$$

and multiplying by $\;r\;$ gives you the result.

Observe though that the imaginary part of the logarithm is $\;\pi\;$ ...without the $\;i\;$ . Both the real and imaginary parts of a complex number are real

  • $\begingroup$ Why is it necessary to multiply by $r$? You have already obtained the result of the original poster. $\endgroup$
    – Angelo
    Feb 7, 2021 at 9:21
  • $\begingroup$ @Angelo That's what was written in the original question. You can check in the edited part. $\endgroup$
    – DonAntonio
    Feb 7, 2021 at 9:52
  • $\begingroup$ I cannot see the original question. What was written in it? $\endgroup$
    – Angelo
    Feb 7, 2021 at 9:55
  • $\begingroup$ @Angelo To the left of the OP's name there's written "edited". If you click there you can see all the versions opf the edited part (in this case, a question). The very first version was to prove that $$r\log\frac{-1+r}r=\pi i$$ which, of course, was wrong as it should have been what is written now: the imaginary part...etc. $\endgroup$
    – DonAntonio
    Feb 7, 2021 at 11:00
  • $\begingroup$ Now, I have just fixed it. Thank you for your clarification. $\endgroup$
    – Angelo
    Feb 7, 2021 at 11:49

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