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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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