How to calculate with imaginary numbers in JavaScript?

Question

Recently, I am trying to calculate using some equations that involve the imaginary number i in them. However, unlike e or π, there isn't any methods or native functions that will return i. A quick search in Google didn't get me any answer. Any ideas on how to achieve it?

function imaginary(){     return {         rational: this,         imaginary: "2i"  //magic code that does this     }; }; Number.prototype.imaginary = imaginary; 

Answer 1

The math.js library supports complex numbers, matrices, and more. The library is compatible with JavaScript's built-in Math library, so quite easy to use.

http://mathjs.org

You can just do things like:

math.i;                         // i math.sqrt(-4)                   // 2i var a = math.complex('2 + 3i'); // 2 + 3i var b = math.complex(4, 5);     // 4 + 5i math.add(a, b);                 // 6 + 8i math.multiply(a, b);            // -7 + 22i math.eval('e^(pi*i) + 1');      // ~0 // etc... 

Edit: note that math.js comes with an expression parser, which makes it more convenient to work with complex values and mathematical expressions:

math.eval('(2 + 3i) * (4 + 5i)'); // -7 + 22i 

Answer 2

Assuming you really want complex numbers, and not just the imaginary component:

I would model a complex number just as you would model a 2D point, i.e. a pair of numbers.

Just as a point has x and y components, so a complex number has real and imaginary components. Both components can just be modeled with ordinary numeric types (int, float, etc.)

However, you will need to define new functionality for all of the mathematical operations.

Addition and subtraction of complex numbers works the same way as addition and subtraction of points - add the separate components to each other, don't mix them. For example:

(3+2i)+(5+4i) = (8+6i)

Multiplication works just like you learned in algebra when multiplying (a+b)*(c+d) = (ac+ad+bc+bd).

Except now you also have to remember that i*i = -1. So:

(a+bi)*(c+di) = (ac+adi+bci+bdii) = (ac-bd) + (ad+bc)i

For division and exponentiation, see http://en.wikipedia.org/wiki/Complex_number