題目：

定義一個複數類Complex，過載運算子“+”、“=”，“*”，“、”，使之能進行復數的加、減、乘、除。運算子過載函式作為Complex類得成員函式。程式設計序，分別求兩個複數之和、差、積和商。

解答： 

#include <iostream>
#include <cmath>
using namespace std;
class Complex{
public:
    Complex(double r=0,double i=0);
    Complex operator +(const Complex& c);
    Complex operator -(const Complex& c);
    Complex operator *(const Complex& c);
	Complex operator /(const Complex& c);
    void print() const;
private:
    double real, imag;
};
Complex::Complex(double r,double i){
    real=r;   
    imag=i;
}
Complex Complex::operator +(const Complex& c){
    double r=real+c.real;
    double i=imag+c.imag;
    return Complex(r,i);
}
Complex Complex::operator -(const Complex& c){
    double r=real-c.real;
    double i=imag-c.imag;
    return Complex(r,i);
}
Complex Complex::operator *(const Complex& c){
    double r=real * c.real - imag * c.imag;
    double i=real * c.imag + imag * c.real;
    return Complex(r,i);
}
Complex Complex::operator /(const Complex& c){
	double t = pow(c.imag,2) + pow(c.real,2);
    double r=real * c.real + imag * c.imag;
    double i=imag * c.real -real * c.imag;
    return Complex(r/t,i/t);
}
void Complex::print() const{
    cout<<'('<<real<<','<<imag<<')'<<endl;
}

int main()
{
	Complex a(3,2),b(0,3),c;
	c = a + b;
	c.print();
	c = a - b;
	c.print();
	c = a * b;
	c.print();
	c = a / b;
	c.print();
	return 0;
}