Re: question about Euler's equation

On May 29, 6:08 pm, Ross <rmill...@xxxxxxxxxxx> wrote:
On May 29, 2:02 pm, The World Wide Wade <aderamey.a...@xxxxxxxxxxx>
wrote:



In article
<2aecbcdc-6521-4714-9a43-29a37fca5...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

eratosthenes <rehamkcir...@xxxxxxxxx> wrote:
I was wondering if Euler's equation can be used to define the log of
negative numbers.

Here is my line of reasoning:

e^(i*pi) + 1 = 0
e^(i*pi) =- 1
i*pi = ln(-1)

so

ln(-a) = ln(-1*a) = ln(-1) + ln(a) = i*pi + ln(a)

It looks nice, but I have a feeling if this worked someone thought of
it a long time ago.

Patrick

But it's also true that e^(3*i*pi) = -1, so by your reasoning ln(-a) =
ln(-1*a) = ln(-1) + ln(a) = 3*i*pi + ln(a). Hmmm...

It is true that if exp(a)=b, then exp(a+2*k*pi*i)=b for any integer
k. Depending upon your application, you can say the logarithm is
multi-valued or you can define a range of 2pi within which the
imaginary part of the logarithm must lie. The problem with defining a
range is that ln becomes a discontinuous function at the edge of the
range. Eratosthenes formula is a specific case of the fact that if we
represent (complex) z as r*exp(i*theta), r and theta real, then
ln(z)=ln(r)+i*theta+2*k*pi*i

that makes a good deal of sense. Thanks.
.