CODEWITHSUNDEEP
javamathcomplex-numbers
Bumzur
Bumzur

Reputation: 111

Calculating complex numbers with rational exponents

Yesterday I created this piece of code that could calculate z^n, where z is a complex number and n is any positive integer.

--snip--
float real = 0;
float imag = 0;

// d is the power the number is raised to [(x + yi)^d]
for (int n = 0; n <= d; n++) {
  if (n == 0) {
    real += pow(a, d);
  } else { // binomial theorem      
    switch (n % 4) {
      case 1: // i
        imag += bCo(d, n) * pow(a, d - n) * pow(b, n);
        break;
      case 2: // -1
        real -= bCo(d, n) * pow(a, d - n) * pow(b, n);
        break;
      case 3: // -i
        imag -= bCo(d, n) * pow(a, d - n) * pow(b, n);
        break;
      case 0: // 1
        real += bCo(d, n) * pow(a, d - n) * pow(b, n);
        break;
    }
  }
}
--snip--

int factorial(int n) {
  int total = 1;
  for (int i = n; i > 1; i--) { total *= i; }
  return total;
}

// binomial cofactor
float bCo(int n, int k) {
  return (factorial(n)/(factorial(k) * factorial(n - k)));
}

I use the binomial theorem to expand z^n, and know whether to treat each term as a real or imaginary number depending on the power of the imaginary number.

What I want to do is to be able to calculate z^n, where n is any positive real number (fractions). I know the binomial theorem can be used for powers that aren't whole numbers, but I'm not really sure how to handle the complex numbers. Because i^0.1 has a real and imaginary component I can't just sort it into a real or imaginary variable, nor do I even know how to program something that could calculate it.

Does anyone know of an algorithm that can help me accomplish this, or maybe even a better way to handle complex numbers that will make this possible?

Oh, I'm using java.

Thanks.

Upvotes: 5

Views: 4727

Answers (4)

Favonius
Favonius

Reputation: 13984

Consider a complex number z such that z = x + iy.

Thus, the polar form of z is = re^itheta, where:

  • r is the magnitude of z, or sqrt(x2+y2), and
  • theta is atan y over x.

Once you have done so, you can use DeMoivre's Theorem to calculate z^n like so:

z^n = r^n e^i n theta

or more simply as

z^n = r^n (cos (n theta) + i sin(n theta))

For more information read up on the polar form of a complex number.

Upvotes: 5

Alexandre C.
Alexandre C.

Reputation: 56976

a^n is ill defined when n is not an integer and a is not a positive number.

If z is a complex number, you can still give a meaning to z^a = exp(a log z) but you have to figure out what log z means when z is not a positive number.

And there is no unique choice.

Upvotes: 0

aioobe
aioobe

Reputation: 421110

First of all, it may have multiple solutions. See Wikipedia: Complex number / exponentiation.

Similar considerations show that we can define rational real powers just as for the reals, so z1/n is the n:th root of z. Roots are not unique, so it is already clear that complex powers are multivalued, thus careful treatment of powers is needed; for example (81/3)4 ≠ 16, as there are three cube roots of 8, so the given expression, often shortened to 84/3, is the simplest possible.

I think you should break it down to polar notation and go from there.

Upvotes: 7

bezmax
bezmax

Reputation: 26142

I'm not really good at math, so probably I understood your task wrong. But as far as I got it - apache commons math can help you: http://commons.apache.org/math/userguide/complex.html

Example:

Complex first  = new Complex(1.0, 3.0);
Complex second = new Complex(2.0, 5.0);

Complex answer = first.log();        // natural logarithm.
        answer = first.cos();        // cosine
        answer = first.pow(second);  // first raised to the power of second

Upvotes: 0

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