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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.

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Bumzur Avatar asked Jun 23 '10 06:06

Bumzur


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1 Answers

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.

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aioobe Avatar answered Sep 19 '22 18:09

aioobe