我有一个名为Complex
的类,它具有实数和虚数的访问器方法,以及将复数表示为字符串的方法和获取复数大小的方法。我在实现最后三个方法 square()
(对复数求平方)、modulusSquared()
(返回复数模数的平方)和最后 >add(Complex d)
,它将把复数 d 添加到这个复数上。我已经尝试过这一点,但我认为我理解错误。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);
}
}
感谢您的宝贵时间。
最佳答案
您的 add
和 square
方法非常简单。据我了解 modulusSquared
是 |x+iy| = Square Root(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);
}
关于java - 如何在 Complex 类中实现 square()、modulus Squared() 和 add() 方法?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35853761/