Re: 2 pi i = 0

[Roger Haar <haar@PHYSICS.ARIZONA.EDU>]

Hi,
        You may be more familiar with the problem with
this "proof" by looking at the sine function.

        sin(0 ) =    0
        sin(2 pi ) = 0

        sin( 0 ) = sin( 2 pi )
        arcsin( sin(0)) = arcsin( sin( 2 pi ))

        0 = 2 pi


        The problem is the arcsin, as used above, is not
single valued.  0 and 2 pi are not on the "same
sheet" as I recall hearing.

        The ln is not singled valued when dealing with
complex numbers.

Thanks
Roger Haar

***********************************************************************
Justin Parke wrote:
> A student showed me a proof of the relationship 2*pi*i=0 this morning.  The proof went:
>
> e^(2*pi*i)=1
>
> e^0=1
>
> e^(2*pi*i)=e^0
>
> take ln of both sides
>
> 2*pi*i = 0
>
> Something seems fishy about this to me.  But I don't know what.
>
> Comments?
>
> This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.