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
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
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 .
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?
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.