My textbook on complex numbers showed this example:
$$ \ln(-\sqrt e) = \ln(-1) + \ln(\sqrt e) = \ln(e^{\pi i}) + \ln(e^{\frac 12}) = \tfrac 12 + \pi i $$
Now, using the same logic, I got this:
$$ \begin{aligned} \ln(5) &= \ln(-1 \cdot -1 \cdot 5) \\\\ &= \ln(-1) + \ln(-1) + \ln(5) \\\\ &= \ln(e^{\pi i}) + \ln(e^{\pi i}) + \ln(5) \\\\ &= \pi i + \pi i + \ln(5) \\\\ &= 2 \pi i + \ln(5) \end{aligned} $$
Subtract $ \ln(5) $ from both sides, and get:
$$ 2 \pi i = 0 $$
Which of course isn't true. I realise that when rotating, $ 2 \pi i $ radians is the same as $ 0 $ radians, but that doesn't explain to me why $ 2 \pi i $ would always be the same as $ 0 $.
I would like to hear by which step my 'proof' went wrong. Thank you!