c语言，求复数之积。利用结构变量求解如下两组复数之积。 za={3,4}, zb={5,6} za={10,20}, zb={30,40} **输出格式要求："(%d+%di)*(%d+%di)=" "(%d+%di)\n" 程序运行示例如下： (3+4i)*(5+6i)=(-9+38i) (10+20i)*(30+40i)=(-500+1000i)

以下是求解两组复数之积的 C 语言代码：

#include <stdio.h>

struct Complex {
    int real;
    int imag;
};

int main() {
    struct Complex za1 = {3, 4};
    struct Complex zb1 = {5, 6};
    struct Complex za2 = {10, 20};
    struct Complex zb2 = {30, 40};

    // 计算第一组复数之积
    int real1 = za1.real * zb1.real - za1.imag * zb1.imag;
    int imag1 = za1.real * zb1.imag + za1.imag * zb1.real;
    printf("(%d+%di)*(%d+%di)=(%d+%di)\n", za1.real, za1.imag, zb1.real, zb1.imag, real1, imag1);

    // 计算第二组复数之积
    int real2 = za2.real * zb2.real - za2.imag * zb2.imag;
    int imag2 = za2.real * zb2.imag + za2.imag * zb2.real;
    printf("(%d+%di)*(%d+%di)=(%d+%di)\n", za2.real, za2.imag, zb2.real, zb2.imag, real2, imag2);

    return 0;
}

输出结果为：

(3+4i)*(5+6i)=(-9+38i)
(10+20i)*(30+40i)=(-500+1000i)