Famously, we have...

where n is any positive integer. This is of course because and

My question is, from this relation that Euler discovered doesn't it also follow that...

?

Since and ?

And if this is true, that means that , right?

Thanks

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- Jul 30th 2010, 12:51 AMrainerthe i*pi=ln(-1) identity
Famously, we have...

where n is any positive integer. This is of course because and

My question is, from this relation that Euler discovered doesn't it also follow that...

?

Since and ?

And if this is true, that means that , right?

Thanks - Jul 30th 2010, 01:08 AMProve It
- Jul 30th 2010, 01:22 AMrainer
- Jul 30th 2010, 01:37 AMHallsofIvy
You cannot go from to so quickly. In the complex numbers the logarithm is multi-valued. Specifically, if, in polar form, then it is also true that . where n can be any integer. In particular, since 1, in polar form, has r= 1, , ln(1) has many values, one of which is 0. The others are, just as you say, multiples of . It does not follow that, just because as a real function, ln(1)= 0 that .

- Jul 30th 2010, 09:18 AMrainer
- Jul 31st 2010, 01:21 AMHallsofIvy
Sorry, I shifted too quickly to the specific case with .

In general, if then it is also true that so

In particular, if x= 1, then r= 1, so for all n. - Jul 31st 2010, 11:32 PMrainer
Wait, if x=1, and consequently r=1 and theta=0, then what you get is...

not as HallsofIvy wrote...

In either case this seems to confirm rather than repudiate the original absurd result I started the thread with via my variation of Euler's identity, viz.:

Correct me if I'm wrong, but as far as I can see we're back to square one again. Any other suggestions as to where the fallacy lies in this absurd result? - Aug 1st 2010, 04:26 AMHallsofIvy
Yes, you are wrong for exactly the reasons stated before: ln(z) is NOT a single valued function. You cannot just assert that ln(1)= 0 in the complex numbers. for any integer m. Yes, m= 0 gives ln(1)= 0 as

**one**of the values of ln(1) but you cannot, in the complex numbers, start from a= ln(1) and declare that a= 0. If a= ln(1) then a is equal to one of the infinite number of possible values of ln(1).

Saying that "[itex]2n\pi i= ln(x)[/itex]" only says that [itex]2n\pi i= 2m\pi i[/itex] for some integer m- which is trivially true.