3 回答

TA貢獻(xiàn)1942條經(jīng)驗(yàn) 獲得超3個(gè)贊
這段代碼對(duì)你有幫助,而且相當(dāng)不言自明:
#include <stdio.h> /* Standard Library of Input and Output */
#include <complex.h> /* Standard Library of Complex Numbers */
int main() {
double complex z1 = 1.0 + 3.0 * I;
double complex z2 = 1.0 - 4.0 * I;
printf("Working with complex numbers:\n\v");
printf("Starting values: Z1 = %.2f + %.2fi\tZ2 = %.2f %+.2fi\n", creal(z1), cimag(z1), creal(z2), cimag(z2));
double complex sum = z1 + z2;
printf("The sum: Z1 + Z2 = %.2f %+.2fi\n", creal(sum), cimag(sum));
double complex difference = z1 - z2;
printf("The difference: Z1 - Z2 = %.2f %+.2fi\n", creal(difference), cimag(difference));
double complex product = z1 * z2;
printf("The product: Z1 x Z2 = %.2f %+.2fi\n", creal(product), cimag(product));
double complex quotient = z1 / z2;
printf("The quotient: Z1 / Z2 = %.2f %+.2fi\n", creal(quotient), cimag(quotient));
double complex conjugate = conj(z1);
printf("The conjugate of Z1 = %.2f %+.2fi\n", creal(conjugate), cimag(conjugate));
return 0;
}
有:
creal(z1):得到真正的部分(浮動(dòng)crealf(z1),長(zhǎng)雙creall(z1))
cimag(z1):得到虛部(浮動(dòng)cimagf(z1),長(zhǎng)雙cimagl(z1))
另外要記住很重要的一點(diǎn),當(dāng)使用復(fù)數(shù)的工作是一樣的功能cos(),exp()并且sqrt()必須與他們復(fù)雜的形式,例如更換ccos(),cexp(),csqrt()。

TA貢獻(xiàn)1831條經(jīng)驗(yàn) 獲得超9個(gè)贊
Complex.h
#include <stdio.h> /* Standard Library of Input and Output */
#include <complex.h> /* Standart Library of Complex Numbers */
int main()
{
double complex z1 = 1.0 + 3.0 * I;
double complex z2 = 1.0 - 4.0 * I;
printf("Working with complex numbers:\n\v");
printf("Starting values: Z1 = %.2f + %.2fi\tZ2 = %.2f %+.2fi\n",
creal(z1),
cimag(z1),
creal(z2),
cimag(z2));
double complex sum = z1 + z2;
printf("The sum: Z1 + Z2 = %.2f %+.2fi\n", creal(sum), cimag(sum));
}
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