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Sorry David
- I didn't see your reply before now.
David Fontaine wrote:
>
> ...
> Speaking of which, I just figured out that to use complex bases and
> exponents you can convert the rectangular vector (a+bi) to circular
> (ai^b)
Hmmm... I think it would be more correct to write it like this:
a + b*i = radius*e^(angle*i)
(The latter is called polar form)
> and then you only have one term and it becomes (somewhat) doable!
What becomes doable ?
> Mind you that I have never had this sort of material in school, I
> conjecture it playing with my TI-89. :) So now I know that n^i =
> cos(ln(n))+i*sin(ln(n)), but I have no idea why. :)
Does this mean that also the TI calculators are
capable of doing symbolic calculations now ?
Below is my try at an explanation of why you get
these results.
If you know Euler's formula:
e^(theta*i) = cos(theta) + i*sin(theta)
And that n can be written as e^ln(n)
and also that (a^b)^c = a^(b*c)
then you'll see that
n^i = (e^ln(n))^i = e^(ln(n)*i) = cos(ln(n)) + i*sin(ln(n))
Euler's formula shows one interesting relation
between these 3 constants: 1, e and i
-e^(pi*i) = 1
(try to feed it to your TI-89 or to a HP-48)
If you want to know more about complex numbers,
then here's a link to a page about the basics:
http://www.math.fsc.qut.edu.au/~gustafso/mab112/week4/
Look at the bottom of the page for info about the
exponential notation of complex numbers.
Also note the 9. property of the exponential form
of a complex number:
That e^(theta*i) is periodic of period 2*pi.
The 10. properties are also noteworthy and often used:
sin(theta) = (e^(theta*i) - e^(-theta*i))/2/i
cos(theta) = (e^(theta*i) + e^(-theta*i))/2
Best regards,
Tor Olav
--
mailto:tor### [at] hotmail com
http://www.crosswinds.net/~tok/tokrays.html
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