Magnitude of complex numbers

Question

I'm working on a project that deals with complex numbers, to explain more (a + bi) where "a" is the real part of the complex number and "b" is the imaginary part of it. (a and b are real numbers but "i" isn't)

I'm having a hard time with implementing a magnitude method, instead of getting back a real number as a double i get infinity.

This is what I'm trying to implement: | (a + bi) | = √(a^2 + b^2)

a snippet off my code

public double mag(){
    /**
     * Compute the magnitude of this complex number.
     * 
     */
    //final double re is the real part
    //final double I'm is the imaginary part 
    double mag = Math.sqrt(Math.pow(this.re, 2) + Math.pow(this.im, 2));

    return mag;
}

say, if both real and imaginary numbers were Double.MAX_VALUE. why would mag return infinity instead of 1.8961503816218352E154 ?

Answer 1

Infinity numbers are numbers larger than Double.MAX_VALUE or smaller than -Double.MAX_VALUE. 1.0/0.0 is infinite. So is 2*Double.MAX_VALUE.

Math.pow(Double.MAX_VALUE,2) is obviously larger than Double.MAX_VALUE. Squaring of a largest possible number will result into something that regular number system can't represent.

Math.sqrt(Math.pow(this.re, 2) + Math.pow(this.im, 2))
==> Math.sqrt(Math.pow(Double.MAX_VALUE, 2) + Math.pow(Double.MAX_VALUE, 2))
==> Math.sqrt(Infinity + Infinity)
==> Infinity

Before the sqrt can be calculated, the squaring of a double max value resulted into infinity. you can't find sqrt of an Infinity number, so what you got it perfectly valid.

Answer 2

for double values if the value exceeds Double.Max_value then it returns infinity. As the other answer has shown so in your case it will return infinity. So for very big numbers you cannot do simple math operations. You better use BigDecimal for this kind of operation.

Answer 3

In order to return 1.8961503816218352E154 do this

    public double mag() {

     double x = this.re;
     double y = this.im;
     return Math.hypot(x, y);

}