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It is known that exists formula for geting a square root of complex number without use of De Moivre formula. Will be interesting if we can find the cubic roots of complex numbers without using De Moivre formula.

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It is not that widely taught that one is pretty much stuck when taking the cube root of a complex number. The theorem is called the Casus Irreducibilis

If an irreducible cubic, with rational coefficients, has three real (and thereby irrational) roots, these may only be found by taking the cube roots of complex numbers, which is what one gets with Cardano's formula. Cardano gives the real roots, in brief, by taking the cube root of a complex number and adding the cube root of its complex conjugate. If one insists on separately finding the real and imaginary parts of the cube root of a complex number, one is led to finding roots of cubics with three real roots. So the whole thing is a bit circular, and in the end we must keep Cardano's description, no further "simplification" is possible.

In conclusion, for cube roots of complex numbers, there is not really anything "better" than De Moivre.

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    $\begingroup$ Put another way: Cardano really isn't that useful in the casus irreduciblis. The more "practical" expressions are the ones that express the three roots in terms of trigonometric or hyperbolic functions. $\endgroup$ – J. M. is a poor mathematician Dec 2 '11 at 0:24
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    $\begingroup$ Good, the real answer. $\endgroup$ – André Nicolas Dec 2 '11 at 0:26
  • $\begingroup$ Something related... $\endgroup$ – J. M. is a poor mathematician Dec 2 '11 at 1:12
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The formula for square roots comes from saying simple that if $\alpha + \beta i = (x + iy)^2$, then you can separate imaginary and reals and solve, more or less. If you want more, I can direct you to a write-up I did recently that happened to include that in it.

Further, it is a (not-so-)good exercise to solve for $x$ and $y$ in $\alpha + \beta i = (x + iy)^3$. There will be a time when it looks like there are more than 3 answers, but you must consider the sign of $\alpha\beta$, which is either positive or negative (this is how you limit solutions in the quadratic case, too).

I say it's not so good because it's painful. Cardano's formula is painful. But I suspect you can convince yourself that it can be done by playing with it for a bit.

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