Complex Number Logarithm Paradox

Asked Mar 08 '19 at 01:31

Active Mar 08 '19 at 04:32

Viewed 69 times

I think I may have stumbled upon a paradox today:

$$\ln(1) = 0$$ $$\ln(1) = \ln(-1^{2}) = 2\ln(-1)$$ $$\ln(-1) = \ln(e^{\pi i}) = \pi i$$ $$2\pi i = 0 \rightarrow i = 0$$

I can't seem to find the error with my logic. Any help would be greatly appreciated

edited Mar 08 '19 at 03:27

rash

asked Mar 08 '19 at 01:31

2

ln(-1) is not uniquely defined – J. W. Tanner Mar 08 '19 at 01:37
1

[Basically what was said in another comment.] The logarithm is multivalued for domains containing more than positive real numbers. – Steve Kass Mar 08 '19 at 01:39
1

You can see the "paradox"/fallacy even more clearly with: $\ln(-1)=\ln(e^{\pi i})=\pi i$ and $\ln(-1)=\ln(e^{-\pi i})=-\pi i$; or even $0=\ln(1)=\ln(e^{2\pi i})=2\pi i$. – Greg Martin Mar 08 '19 at 02:02
The error is in the last step: $2\pi i=0 \to \pi=0$ is correct. – vadim123 Mar 08 '19 at 04:42
@vadim123 how is that correct – rash Mar 08 '19 at 05:10
@rash, it is just as correct as the OP's reasoning, and meant to humorously illustrate the flaw. – vadim123 Mar 08 '19 at 14:23

0 Answers0