Trigonometry Problem without Trigonometry: Clarification/Correction

In the previous post I was exulting about the coolness of solving a trigonometry problem without trigonometry, using complex numbers. Well, that isn’t quite the full truth. You can certainly do the problem described easily and elegantly with a complex number capable calculator. However, there is some hidden trigonometry in the initial rotation operation:

The complex number eⁱ has to be raised to a power. If you raise to an integer power, you could simply multiply out the rectangular values algebraically. But how do you raise a complex number to a fractional power, like 2.5? The standard approach, near as I can tell, is to convert the complex number to polar format, with a magnitude r and an angle ϕ. Assuming complex number a+bi, pulling out the magnitude requires only the sqrt function (pythagorean theorem) but pulling out the angle requires the arctangent of b/a. Then, use the formula Zⁿ = rⁿ (cos nϕ + i sin nϕ). I really don’t know if this is how all calculators do it, but here is the code behind the operations in the racket interpreter:

  Scheme_Complex *cb = (Scheme_Complex *)base;
  Scheme_Complex *ce = (Scheme_Complex *)exponent;
  double a, b, c, d, bm, ba, nm, na, r1, r2;
  int d_is_zero;

  if ((ce->i == zero) && !SCHEME_FLOATP(ce->r)) {
    if (SCHEME_INTP(ce->r) || SCHEME_BIGNUMP(ce->r))
      return scheme_generic_integer_power(base, ce->r);
  }

  a = scheme_get_val_as_double(cb->r);
  b = scheme_get_val_as_double(cb->i);
  c = scheme_get_val_as_double(ce->r);
  d = scheme_get_val_as_double(ce->i);
  d_is_zero = (ce->i == zero);

  bm = sqrt(a * a + b * b);
  ba = atan2(b, a);

  /* New mag & angle */
  nm = scheme_double_expt(bm, c) * exp(-(ba * d));
  if (d_is_zero) /* precision here can avoid NaNs */
    na = ba * c;
  else
    na = log(bm) * d + ba * c;

  r1 = nm * cos(na);
  r2 = nm * sin(na);

The code there is just slightly more complicated, as it has separate angle finding conditional branches for real number and complex number exponents.

Of course, you don’t need to deal with any of that if you are just manipulating complex numbers in the interpreter or calculator.

I feel this would be a great place to dive into the interesting subject of complex roots, but that I should hold off until I have a more thorough grasp of the subject.