Question

我有一个名为Complex的类，它具有实数和虚数的存取方法，以及将复数表示为字符串的方法和获取复数的大小的方法。我无法实现我的最后三个方法square()，它将复数modulusSquared()平方，它返回复数模数的平方，最后返回add(Complex d)，这将添加复数d到这个复数。我试过这个，但我想我理解这个错了。这是我的代码：

public class Complex { //real + imaginary*i

    private double re;
    private double im;

    public Complex(double real, double imaginary) {
        this.re = real;
        this.im = imaginary;
    }

    public String toString() { //display complex number as string
        StringBuffer sb = new StringBuffer().append(re);
        if (im>0)
          sb.append('+');
        return sb.append(im).append('i').toString();
      }

    public double magnitude() {   //return magnitude of complex number
        return Math.sqrt(re*re + im*im);
      }

    public void setReal(double m) {
        this.re = m;
    }

    public double getReal() {
        return re;
    }

    public void setImaginary(double n) {
        this.im = n;
    }

    public double getImaginary() {
        return im;
    }

    public void square() {    //squares complex number
        Complex = (re + im)*(re + im);
    }

    public void modulusSquared() {   //returns square of modulus of complex number
        Math.abs(Complex);
    }

    public void add(Complex d) {    //adds complex number d to this number

        return add(this, d);

    }

}

感谢您的时间。

Answer 1

您的add和square方法非常简单。据我所知，modulusSquared是 | x + iy | =平方根（x ^ 2 + y ^ 2），因此实现也非常简单：

public void add(Complex d) {    // adds complex number d to this number
    this.re += d.getReal();     // add real part
    this.im += d.getImaginary();// add imaginary part
}
public void square(){
    double temp1 = this.re * this.re; // real * real
    double temp2 = this.re * this.im; // real * imaginary (will be the only one to carry around an 'i' wth it
    double temp3 = -1 * this.im * this.im; // squared imaginary so multiply by -1 (i squared)
    this.re = temp1 + temp3; // do the math for real
    this.im = temp2;         // do the math for imaginary
}

public double modulusSquared() { // gets the modulus squared
    return Math.sqrt(this.re * this.re + this.im * this.im); 
}

如何在Complex类中实现square（），modulusSqaured（）和add（）方法？

1 个答案: