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Is there a general equation for a cube root of a complex number that does not exploit De Moivre's Theorem or in any way use trigonometric functions?

For example, a square root of a complex number $x$ is $$\sqrt{\frac{|x|+\operatorname{Re}(x)}{2}}+i\sqrt{\frac{|x|-\operatorname{Re}(x)}{2}}.$$ Is there a similar equation for a cube root of $x$?

By introducing $a$ and $b$ such that $(a+ib)^3=x$, we can then expand and obtain two equations in $a$ and $b$ by equating the real and imaginary parts on each side.

However, the resulting equations are cubic and I don't know any method to find the roots of a cubic equation without having to take the cube root of a complex number, which is the problem I want to solve in the first place.

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  • $\begingroup$ The imaginary part should have $\text{sgn}(\text{Im}(x))$; otherwise the square is $\text{Re}(x)+i|\text{Im}(x)|$. $\endgroup$
    – mr_e_man
    Commented Feb 17, 2020 at 20:43
  • $\begingroup$ I just searched for this while editing this answer. For the "Trisecting an angle" section, I need the cube root of a complex number. But I can't use De Moivre's Theorem because the whole point of the exercise is to find expressions for $\cos(\theta/3)$ and $\sin(\theta/3)$. $\endgroup$
    – Dan
    Commented Aug 2, 2022 at 16:01

2 Answers 2

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There is no such nice formula for the cube root of a complex number with both real and imaginary parts nonzero. If you write out the real and imaginary parts of your cube root, you wind up solving cubic equations in one variable that have three irrational roots. This is the Casus Irreducibilis http://en.wikipedia.org/wiki/Casus_irreducibilis

In turn, for each of those cubics, Cardano's method leads you to finding the cube roots of other complex numbers. All very circular, and never gets anywhere.

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  • $\begingroup$ I wonder: Is this in any way related to the general impossibility of trisecting angles in classical geometric constructions? $\endgroup$
    – Brian Tung
    Commented Aug 18, 2015 at 23:57
  • $\begingroup$ @BrianTung: No; trisection is impossible by straightedge and compass for a different (and much simpler) reason: each such construction will produce a coordinate in a quadratic extension (of the previous field) if at all, and hence every such constructible point lies in an extension of $\mathbb{Q}$ of degree a power of $2$, and by the tower law must have minimal polynomial with degree also a power of $2$. $\cos(20^\circ)$ has a cubic minimal polynomial over $\mathbb{Q}$, and hence cannot be constructible in this way. But the neusis construction can do general trisection. $\endgroup$
    – user21820
    Commented Nov 27, 2016 at 13:59
  • $\begingroup$ @BrianTung: Anyway I meant the reason for the impossibilities are not related, and they are related only through the fact that casus irreducibilis is equivalent to irreducible rational cubic with 3 real roots, which is in turn related to the constructibility of cosines (via irreducibility) since the cubic $( 4x^3 - 3x - \cos(t) )$ has 3 real roots (via intermediate value theorem; try $x \in \{-1,-\frac12,0,1\}$). There is no connection with the original definition of casus irreducibilis in terms of radicals. $\endgroup$
    – user21820
    Commented Nov 27, 2016 at 14:13
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$$\sqrt[3]{u+iv}=w+ix$$

$$\boxed{\boxed{\sqrt[3]{u}\sum_{k\geqslant 0}\binom{\frac{1}{3}}{2k}(-1)^{k}\left(\frac{v}{u}\right)^{2k}=w\\\frac{v\sqrt[3]{u}}{u}\sum_{k\geqslant 0}\binom{\frac{1}{3}}{2k+1}(-1)^{k}\left(\frac{v}{u}\right)^{2k}=x}}$$

Inversely,

$$\boxed{u=w^3-3wx^2\\v=3w^2x-x^3}$$

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