z=(8+6i)(4+3i)/(8+6i)+(4+3i)，把上述抽象数据举例利用Dev-c++语言进行代码输出

在C++中，由于复数（complex numbers）不是内置的数据类型，我们需要自定义一个`Complex`类来处理复数运算。对于给定的表达式 `z = (8+6i) * (4+3i) / (8+6i) + (4+3i)`，我们可以创建一个简单的实现来计算它。假设已经有一个名为`Complex`的简单复数类，代码会类似下面这样：

#include <iostream>
using namespace std;

class Complex {
public:
    double real;
    double imag;

    // 构造函数
    Complex(double r = 0, double i = 0) : real(r), imag(i) {}

    // 复数乘法
    Complex operator*(const Complex& other) {
        return Complex(real * other.real - imag * other.imag,
                      real * other.imag + imag * other.real);
    }

    // 复数除法
    Complex operator/(const Complex& other) const {
        double denominator = other.real * other.real + other.imag * other.imag;
        return Complex((real * other.real + imag * other.imag) / denominator,
                      (imag * other.real - real * other.imag) / denominator);
    }
};

int main() {
    Complex num1(8, 6);  // (8+6i)
    Complex num2(4, 3);  // (4+3i)

    // 计算乘法
    Complex product = num1 * num2;
    cout << "Product: (" << product.real << ", " << product.imag << ")\n";

    // 除以num1，注意这里num1不会分母为零，因为分母是一个非零复数
    Complex quotient = product / num1;
    cout << "Quotient (after removing complex division): (" << quotient.real << ", " << quotient.imag << ")\n";

    // 后面两个加法操作不需要特殊处理，直接相加即可
    Complex sum = quotient + num2;  // z = quotient + (4+3i)
    cout << "Sum: (" << sum.real << ", " << sum.imag << ")\n";

    return 0;
}

运行这段代码后，你会看到结果的实部和虚部分别是什么。