#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 123 4311
"Lane Straatman" <in*****@invali d.netwrote in message
>
Why doesn't e^(i *pi) equal what most folks think it does? LS
Most folks would say that if you try to multiply a number by itself an
imaginary number of times, that is impossible.
In article <44************ *************** ***@comcast.com >,
Lane Straatman <in*****@invali d.netwrote:
[much snippage]
>z5= 0 + I*(3.14159);
z3=2.54 + 0*I;
If I understand your post correctly, this should be
z3=2.7182818284 59045 + 0*I;
>z4=cpow(z5,z3) ; printf("%lf %lf\n", creal(z4), cimag(z4));
>Why doesn't e^(i *pi) equal what most folks think it does? LS
'Tmight come closer if e were what most people think it is.
dave
--
Dave Vandervies dj******@csclub .uwaterloo.ca
This means that, if I break into a slowish run, then according to the
Lorentz-Fitzgerald equations I acquire imaginary mass, and can only decelerate
by running harder. --Richard Heathfield in comp.programmin g
"Malcolm McLean" <re*******@btin ternet.comwrote in message
news:9O******** *************** *******@bt.com. ..
>
"Lane Straatman" <in*****@invali d.netwrote in message
>> Why doesn't e^(i *pi) equal what most folks think it does? LS
Most folks would say that if you try to multiply a number by itself an
imaginary number of times, that is impossible.
e is the first transcendental (Louisville, I believe). pi is the second.
They are not equal. LS
Malcolm McLean wrote:
"Lane Straatman" <in*****@invali d.netwrote in message
>> Why doesn't e^(i *pi) equal what most folks think it does?
Most folks would say that if you try to multiply a number by
itself an imaginary number of times, that is impossible.
Not in the complex field.
--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net>
Lane Straatman wrote:
"Malcolm McLean" <re*******@btin ternet.comwrote in message
>"Lane Straatman" <in*****@invali d.netwrote in message
>>> Why doesn't e^(i *pi) equal what most folks think it does?
Most folks would say that if you try to multiply a number by itself an imaginary number of times, that is impossible.
e is the first transcendental (Louisville, I believe). pi is
the second. They are not equal.
No such thing as 'first' or 'second' transcendental. Between any
two rational numbers, there are an infinity of transcendentals .
Look up the various alephs. Also Cantor, Dedekind, Wierstrass. C
floating point representation ignores this fundamental fact and
represents all values by a limited set of rationals. Note the sly
maintenance of topicality.
--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net>
Malcolm McLean wrote:
"Lane Straatman" <in*****@invali d.netwrote in message
>Why doesn't e^(i *pi) equal what most folks think it does? LS
Most folks would say that if you try to multiply a number by itself an
imaginary number of times, that is impossible.
That's plane wrong.
--
Eric Sosman es*****@acm-dot.org.invalid
CBFalconer said:
Malcolm McLean wrote:
>"Lane Straatman" <in*****@invali d.netwrote in message
>>> Why doesn't e^(i *pi) equal what most folks think it does?
Most folks would say that if you try to multiply a number by itself an imaginary number of times, that is impossible.
Not in the complex field.
Read again, more carefully. Malcolm is not talking about imaginary
arithmetic, but about what "most folks would say" about it. Most folks
don't know spit about mathematics.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999 http://www.cpax.org.uk
email: rjh at the above domain, - www.
Eric Sosman said:
Malcolm McLean wrote:
>"Lane Straatman" <in*****@invali d.netwrote in message
>>Why doesn't e^(i *pi) equal what most folks think it does? LS
Most folks would say that if you try to multiply a number by itself an imaginary number of times, that is impossible.
That's plane wrong.
Either you have a couple of axes to grind, or you're misreading what Malcolm
wrote, which rings true.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999 http://www.cpax.org.uk
email: rjh at the above domain, - www.
"Richard Heathfield" <rj*@see.sig.in validwrote in message
CBFalconer said:
>Malcolm McLean wrote:
>>"Lane Straatman" <in*****@invali d.netwrote in message
Why doesn't e^(i *pi) equal what most folks think it does?
Most folks would say that if you try to multiply a number by itself an imaginary number of times, that is impossible.
Not in the complex field.
Read again, more carefully. Malcolm is not talking about imaginary
arithmetic, but about what "most folks would say" about it. Most folks
don't know spit about mathematics.
If we multiply an imaginary number by itself an imaginary number of times,
maybe that would be real, on the analogy that a negative number multiplied
by itself is a positive.
Let's try it
int main(void)
{
int unity = 1;
int test1;
test1 = 0 - unity;
printf("%d * %d = %d\n", test1, test1, test1 * test1);
printf("Now a bit more complex %f\n", pow( sqrt(test1), sqrt(test1) );
return 0;
}
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