Re: complex cube roots
- From: "barr" <barr@xxxxxxxxx>
- Date: 31 Oct 2006 04:41:12 -0800
Geoff Harland wrote:
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.)
"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_. "
I was afraid that that was the case. I did know that you cannot find
the real root(s) of general cubics without going into C, but I had
hoped that perhaps the simpler question of finding a complex cube root
might be tractable. Although now I realize that that would resolve the
case of a real cubic with exactly one real root, since that requires
that the discriminant be negative and then you have to find a square
root of it and then a cube root of a real added to it. This involves
complex numbers of course, but if you could find formulas for the real
and imaginary parts of a complex cube root, you could write a formula
in which complex numbers did not appear.
.
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