|
|
barr wrote:
> It is well known and easy to prove that square roots of complex numbers
> can be calculated using real square roots. That is you can solve the
> equation (x + iy)^2 = a + ib for x and y using only real square roots.
> My question is, is it possible to calculate complex cube roots using
> only real cube and, if necessay, square roots. I tried, but the
> equations I got seemed intractible.
Been there, done that (and got the tee-shirt). :-)
I can't recall who proved it, or when, but I definitely do remember reading
that it has been proved that it is not possible to define the real part (or
the imaginary part) of the cube root of a "general" complex number using a
_finite_ number of radicals (square roots, cube roots, etc) if _all_ of the
associated numbers are (to be) _real_.
(After a bit of dabbling on one occasion, I suspected that that was the
case, and not that long afterwards, determined, after reading a book ("The
History of Mathematics"?), that my hunch was in fact correct.)
You possibly know that trigonometric methods can be used to determine the
cube roots of complex numbers, but "in general" that technique won't give
you the outcome you are looking for (and in any particular cases when cube
roots _can_ be expressed in "real-only radicals", those roots could always
be determined _without_ resorting to trigonometric methods).
Regards,
Geoff Harland.
g_harland@xxxxxxxxxxxxxxxxx
(Transpose m & s in address
provided - then also remove
cuberoot of 10^3 + 9^3 - 1^3.)
|
|