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

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 $$\pi i$$.

• What is the meaning $r$ in the both side ? Also $r=0$ is not possible Feb 7, 2021 at 8:03
• 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. Feb 7, 2021 at 8:06
• My bad, I will edit the question
– runr
Feb 7, 2021 at 8:08
• I think it should be $\;r\in(0,1)\;$ , as in the extreme points of the unit interval the expression isn't defined Feb 7, 2021 at 8:16
• @DonAntonio you're right, thanks a lot!
– runr
Feb 7, 2021 at 8:17

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

$$\log\frac{-1+r}r=\ln\left|\frac{-1+r}r\right|+i\arg\frac{-1+r}r$$

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

$$0

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

• Why is it necessary to multiply by $r$? You have already obtained the result of the original poster. Feb 7, 2021 at 9:21
• @Angelo That's what was written in the original question. You can check in the edited part. Feb 7, 2021 at 9:52
• I cannot see the original question. What was written in it? Feb 7, 2021 at 9:55
• @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. Feb 7, 2021 at 11:00
• Now, I have just fixed it. Thank you for your clarification. Feb 7, 2021 at 11:49