[seqfan] Re: Does (e i)^(pi i) have an imaginary part?
Andrew N W Hone
A.N.W.Hone at kent.ac.uk
Tue Aug 10 14:53:48 CEST 2010
Indeed you are right. I was being rather cavalier in working only with the principal branch of the logarith (n=0), in order
to recover the output from Mathematica, and had not noticed that you had already given a much more accurate account
in an earlier message.
From: seqfan-bounces at list.seqfan.eu [seqfan-bounces at list.seqfan.eu] On Behalf Of Robert Israel [israel at math.ubc.ca]
Sent: 08 August 2010 18:39
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Does (e i)^(pi i) have an imaginary part?
On Sun, 8 Aug 2010, Andrew N W Hone wrote:
> Hi -
> It's best to learn how to unravel this without a computer if you can, as with mathematica or maple
> one can always worry about whether zero really means zero!
> I would suggest reading an introductory text on complex analysis: the complex exponential is usually one of
> the first examples of analytic functions in any such book.
> The multi-valuedness of the logarithm may take a bit longer to understand than the exponential.
> However, if you know how the ordinary (real) exp and log functions work, then you should quickly get the hang of the
> complex case.
>> From the basic rules of exponentiation,
> (a b)^c = a^c b^c
Stop right there. This is not valid in general for complex numbers.
The basic rules are
a^b = exp(b log a)
exp(a + b) = exp(a) exp(b)
if exp(a) = exp(b) then a - b = 2 pi n i for some integer n
So (a b)^c = exp(c log(a b))
while a^c b^c = exp(c (log a + log b))
and log(a b) = log a + log b + 2 pi n i
so (a b)^c = exp(2 pi n c i) a^c b^c
> means that
> (e i)^(pi i) = e^(pi i) i^(pi i) = - i^(pi i).
> Also, exp and log are inverses of each other, and one of the basic rules of logs is that
> log (a^b) = b log a,
Again not true in general: log(a^b) = b log a + 2 n pi i.
Robert Israel israel at math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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