[seqfan] Re: Does (e i)^(pi i) have an imaginary part?

[Robert Israel]

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