c语言实现：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)

以下是求解复数之积的 C 语言代码：

#include <stdio.h>

struct complex {
    int real;
    int imag;
};

int main() {
    struct complex za = {3, 4};
    struct complex zb = {5, 6};
    struct complex zc = {10, 20};
    struct complex zd = {30, 40};

    // 计算第一组复数之积
    struct complex res1 = {0, 0};
    res1.real = za.real * zb.real - za.imag * zb.imag;
    res1.imag = za.real * zb.imag + za.imag * zb.real;
    printf("(%d+%di)(%d+%di)= (%d+%di)\n", za.real, za.imag, zb.real, zb.imag, res1.real, res1.imag);

    // 计算第二组复数之积
    struct complex res2 = {0, 0};
    res2.real = zc.real * zd.real - zc.imag * zd.imag;
    res2.imag = zc.real * zd.imag + zc.imag * zd.real;
    printf("(%d+%di)(%d+%di)= (%d+%di)\n", zc.real, zc.imag, zd.real, zd.imag, res2.real, res2.imag);

    return 0;
}

输出结果为：

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