Re: 2 pi i = 0

[Uri Ganiel <uri.ganiel@WEIZMANN.AC.IL>]

At 10:03 -0500 22/11/02, 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.

Common mistake, yet clearly wrong: e^(i*x) is periodic, with a period 2*pi.
Hence you could infer for any x that: x=x+2*n*pi with n an integer...
Clearly, you could do the same for the trigonometric functions (sine
and cosine).
Uri

--
Uri Ganiel
The Rudy Bruner Professor of Science Teaching
Department of Science Teaching
The Weizmann Institute of Science
Rehovot 76100, ISRAEL

Tel: 972-8-9343895
FAX: 972-8-9473677
E-Mail: uri.ganiel@weizmann.ac.il

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