e to the i pi

[Lane Straatman]

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <complex.h>
/*
double complex z1, z2, z3;
bool flag;
z1 = .4 + .7I;
z2 = cpow(z1, 2.0);
z3 = z1 * z1;
flag = false;
flag = true;
if (flag)
{
printf("%lf %lf\n", creal(z1), cimag(z1));
printf("%lf %lf\n", creal(z2), cimag(z2));
printf("%lf %lf\n", creal(z3), cimag(z3));
printf("%d\n", N);
}
*/

int main(int argc, char *argv[])
{
double complex z1, z2, z3, z4, z5;
z1=5 +7I;
z2=cpow(z1, 1I);
printf("%lf %lf\n", creal(z1), cimag(z1));
printf("%lf %lf\n", creal(z2), cimag(z2));
z5= 0 + I*(3.14159);

z3=2.54 + 0*I;
z4=cpow(z5,z3);
printf("%lf %lf\n", creal(z4), cimag(z4));

system("PAUSE") ;
return 0;
}
Why doesn't e^(i *pi) equal what most folks think it does? LS

[David Wade]

"Malcolm McLean" <re*******@btin ternet.comwrote in message
news:H9******** *************@b t.com...

>
"Dave Vandervies" <dj******@caffe ine.csclub.uwat erloo.cawrote in message

In article <eq***********@ pc-news.cogsci.eds ..ac.uk>,
Richard Tobin <ri*****@cogsci .ed.ac.ukwrote:

>In article <eL************ *************** ***@bt.com>,
Malcolm McLean <re*******@btin ternet.comwrote :

Can't say this bit about e^(i*x) = cos x + i * sin x is too convincing
though.

Consider the polynomial expansion e^x = 1 + x + x^2/2! + x^3/3! + ...

I think you mean "power series expansion". It's only a polynomial if
it's finite.
dave
(knowing the power series expansions for sin and cos would also be
helpful)

You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

For the same reason that the formulas for the roots of quadratic equations
are not self evident. In general, to mathematicians, things only get
interesting when we leave the tangible and touchable behind, and move to a
higher level of abstraction. Trying to relate these back to physical models
in these cases is often unproductive. What you are asking is almost the same
as asking what specific numbers are "x" and "y" in an formula such as
y=x+4...

[Dik T. Winter]

In article <bo************ *************** ***@eclipse.net .uk"David Wade" <g8***@yahoo.co mwrites:

"Malcolm McLean" <re*******@btin ternet.comwrote in message
news:H9******** *************@b t.com...

"Dave Vandervies" <dj******@caffe ine.csclub.uwat erloo.cawrote in message

....

(knowing the power series expansions for sin and cos would also be
helpful)
>

You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

For the same reason that the formulas for the roots of quadratic equations
are not self evident.

No. It is much more basic. How would you define a trangle with one right
angle and two complex angles? The question being: "what *is* a complex
angle?"
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, [phone]
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

[Malcolm McLean]

"Dik T. Winter" <Di********@cwi .nlwrote in message

No. It is much more basic. How would you define a trangle with one right
angle and two complex angles? The question being: "what *is* a complex
angle?"

Let's imagine a triangle with complex instead of real coordinates.

So instead of x1, y1, x2, y2, x3, y3 we have xr1, xi1, yr1, yi1, xr2, xi2,
yr2, yi2, xr3, xi3, yr3, yi3.

You agree that the sine of such a triangle could be complex, if only we can
decide which is the right angle, because we are dividing one complex number
by another.

To reduce thr strain on our brain, lets declare point one to be at the
origin.

Now we have

0, 0, 0, 0, xr2, xi2, yr2, yi2, xr3, xi3, yr3, yi3.

To reduce the strain further, we can say that point two always alies along
the real x axis.

0, 0, 0, 0, xr2, 0, 0, 0, xr3, xi3, yr3, yi3.

Once we've done that we can rotate about the x axis until one of the other
points goes to zero. Let's choose to lose yi3.

Now we've got

0, 0, 0, 0, xr2, 0, 0, 0, xr3, xi3, yr3, 0.

Now things are really looking up. We've only got one imaginary component
complicating things.

However can can play another trick. Three points in four dimensions are
exactly the same as four points in three dimensions.

So now we;ve got

0, xr2, xr3 - xreal
0, 0, xi3 - x imaginary
0, 0, yr3 - y real
0, 0, 0 - y imaginary.

very nicely, we've got a point at the origin. And four points can be defined
by two lengths, two angles, and one torsion angle - assuming we ignore
translation and rotation. However when we look at our zeroes, we can knock
out the x column, and the two middle values are in a straight line. So we
are actually left with a regular triangle, and we can see if it is
right-angled. Unfortunately we actually have three such "triangles" , because
we've introduced more possible connections. Only one of them has no BC edge
and thus isn't a triangle.

[Dik T. Winter]

In article <KM************ *********@bt.co m"Malcolm McLean" <re*******@btin ternet.comwrite s:

"Dik T. Winter" <Di********@cwi .nlwrote in message

No. It is much more basic. How would you define a trangle with one right
angle and two complex angles? The question being: "what *is* a complex
angle?"

Let's imagine a triangle with complex instead of real coordinates.

I ask what a complex angle is. Or more basic: I want to compute the
sine of 2.i. You would be a lot more convincing if you could construct
a triangle with angles pi/2, 2.i and ( pi/2 - 2.i). In that case we may
use the definition of quotients of sides. We will not, of course, because
the modulus of such a quotient will alway be less than or equal to 1, but
sin(2.i) = 3.626860408 .i.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, [phone]
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/