e到i pi [英] e to the i pi
问题描述
#include< stdio.h>
#include< stdlib.h>
#include< stdbool.h>
#include< complex.h>
/ *
双复数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 []）
{
双复数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"）;
返回0;
}
为什么^（i * pi）与大多数人认为的相同？ LS
#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
推荐答案
" Lane Straatman" < in ***** @ invalid.netwrote in message
"Lane Straatman" <in*****@invalid.netwrote in message
>
为什么没有^（i * pi ）大多数人认为它的作用相同？ LS
>
Why doesn''t e^(i *pi) equal what most folks think it does? LS
大多数人会说，如果你试图将一个数字乘以一个
虚数，那是不可能的。
Most folks would say that if you try to multiply a number by itself an
imaginary number of times, that is impossible.
文章< 44 *************************** ***@comcast.com>，
Lane Straatman< in ***** @ invalid.netwrote：
[多次狙击]
In article <44******************************@comcast.com>,
Lane Straatman <in*****@invalid.netwrote:
[much snippage]
> z5 = 0 + I *（3.14159）;
z3 = 2.54 + 0 * I;
>z5= 0 + I*(3.14159);
z3=2.54 + 0*I;
如果我理解你的帖子，那应该是
z3 = 2.718281828459045 + 0 * I;
If I understand your post correctly, this should be
z3=2.718281828459045 + 0*I;
> z4 = cpow（z5，z3）;
printf（"％lf％lf \ n"，creal（z4），cimag（z4））;
>z4=cpow(z5,z3);
printf("%lf %lf\n", creal(z4), cimag(z4));
>为什么e ^（i * pi）与大多数人认为的相同？ LS
>Why doesn''t e^(i *pi) equal what most folks think it does? LS
''如果e是大多数人认为的那样，那么接近。
dave
-
Dave Vandervies dj******@csclub.uwaterloo.ca
这意味着，如果我进入一个缓慢的运行，那么根据洛朗兹 - 菲茨杰拉德方程式我获得虚数质量，并且只能减速<通过更加努力地运行
。 - 理查德程序中的理查德希思菲尔德
''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.programming
" Malcolm McLean" < re ******* @ btinternet.com写信息
新闻：9O *********************** ******* @ bt.com ...
"Malcolm McLean" <re*******@btinternet.comwrote in message
news:9O******************************@bt.com...
>
" Lane Straatman" < in ***** @ invalid.netwrote in message
>
"Lane Straatman" <in*****@invalid.netwrote in message
>>
为什么没有^（i * pi）等于大多数人认为它有什么作用？ LS
>>
Why doesn''t e^(i *pi) equal what most folks think it does? LS
大多数人会说，如果你试图将一个数字乘以一个
的虚数次，这是不可能的。
Most folks would say that if you try to multiply a number by itself an
imaginary number of times, that is impossible.
e是第一个超然（路易斯维尔，我相信）。 pi是第二个。
它们不相等。 LS
e is the first transcendental (Louisville, I believe). pi is the second.
They are not equal. LS
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