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Path: utzoo!linus!decvax!mcnc!akt
From: akt@mcnc.UUCP (Amit Thakur)
Newsgroups: net.math
Subject: Re: References on i ** i, "principal logs"
Message-ID: <1964@mcnc.UUCP>
Date: Sun, 19-Feb-84 13:19:23 EST
Article-I.D.: mcnc.1964
Posted: Sun Feb 19 13:19:23 1984
Date-Received: Mon, 20-Feb-84 01:05:16 EST
References: <205@pucc-i>, <1960@mcnc.UUCP>
Organization: Microelectronics Ctr. of NC; RTP, NC
Lines: 31


  i been doing some more wondering overnight:

 ln(-1) = i*pi(1+2n*pi) because exp{ i*pi(1+2n*pi)} = -1.
 i think this is a better solution.

 also: exp{ +-i*2n*pi } = 1. 0 = ln(1) = +-i*2n*pi
 but 0 <> +-i*2n*pi for n <> 0.
 note that +-i*2n*pi = +-i*2n*pi{exp[i*arctan(+-2n*pi/0)]} =
 +-i*2n*pi{exp[i*arctan(+-infinity)]} = +-i*2n*pi{exp[i*+-pi/2]}
 applying euler's formula gives us +-i*2n*pi again, as it should.

 thus, if we let n=0, then obviously, 0 = +-i*2n*pi, and we
 can then say that ln(0) = 0. or, if we choose any other n,
 and force 0 = +-i*2n*pi, we can say that ln(0) = +-i*2n*pi.
 
 there: ln() is now defined over the entire complex plane!
 question: i know i divided by zero to arrive at the above
 result, but i don't see why this isn't valid, especially since
 the exp() function is defined as an infinite series.
 please: no flames to the effect that i should go back and study
 my high school algebra. tell me why this is not valid in *this*
 particular case.

 of course, using ln(x)/ln(b), one could easily find log (x).
                                                        b
akt at ...decvax!mcnc!akt

p.s. this chain of thoughts was started by the idea that pure
negative imaginary numbers had logs, but not negative reals.
and why should zero be so lonesome all by itself?