Ernie,

I think you're wrong. There really are three cube roots for all numbers - even real ones.

For sure, for negative, real numbers, (such as -2, -3, -4, etc.), there are the regular cube root (i.e. the same cube root as for the equivalent positive number, but with a minus sign in front) as well as those roots rotated by +/-120 degrees in the complex plane. (The Mercedes symbol has to be rotated by -90 degrees for positive numbers, and +90 degrees for negative numbers, where + implies counterclockwise rotation. The "simple" (non-imaginary) root lies along the +x axis for positive numbers and along the -x axis for negative numbers.)

For example: the cube root of +4 returned by the 42S is 1.5874 (using the y^x button, with y=4 and x=1/3), i.e. along the +x (positive real numbers) axis. However, the complex roots -0.794-1.37i and -0.794+1.37i also give +4 when cubed (neglecting round-off, of course).

Conversely, the cube root of -4 determined by the same set of button pushings gives +0.794+1.37i with the 42S. But, -1.5874^3 = -4, as does (0.794-1.37i)^3 . (Again, to within round-off)

Similar results hold for any positive or negative (real; i.e. non-imaginary) number. Thus, Raul had it right (with my minor correction).

The extension to any power is that there are n values for the n-th root, arranged as a Mercedes star with n points equally spaced in angle in the complex plane, with one point aligned with the +x (real) axis for positive numbers, and one point aligned with the -x (also real) axis for negative numbers.

I leave complex values for our mathematician friends, with the expectation that one of the star points of the root system will be either aligned or anti-aligned with the original complex number.

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