# Thread: Real algebraic Expressions for cubic root?

1. ## Real algebraic Expressions for cubic root?

We know the square roots of a complex number $a + ib \quad (a,b \in \mathbf{R})$ can be written as
$x + iy = \pm \frac {1}{\sqrt {2}} (\sqrt {a + \sqrt {a^2 + b^2}} + i sgn(b + ) \sqrt { - a + \sqrt {a^2 + b^2}})$
where $sgn(b + ) = 1$ when $b \geq 0$ otherwise $sgn(b + ) = - 1$

Thus we have real algebraic expressions for the real and imaginary parts of the square root, without trigonometric functions involved.

Does this still the case for cubic roots? I tried with no success. Thought that's not possible but cannot prove my guess

2. Originally Posted by elim
We know the square roots of a complex number $a + ib \quad (a,b \in \mathbf{R})$ can be written as
$x + iy = \pm \frac {1}{\sqrt {2}} (\sqrt {a + \sqrt {a^2 + b^2}} + i sgn(b + ) \sqrt { - a + \sqrt {a^2 + b^2}})$
where $sgn(b + ) = 1$ when $b \geq 0$ otherwise $sgn(b + ) = - 1$

Thus we have real algebraic expressions for the real and imaginary parts of the square root, without trigonometric functions involved.

Does this still the case for cubic roots? I tried with no success. Thought that's not possible but cannot prove my guess
You want to find $p,q\in\mathbb{R}$ such that $\sqrt[3]{a+bi} = p+qi \implies a+bi = (p+qi)^3 \implies p^3-3pq^2+(3p^2q-q^3)i = a+bi$

Therefore we have two equations and two unknowns:

$\begin{cases} p^3-3pq^2 = a\\3p^2q-q^3 = b \end{cases}$

Looks rather messy to solve, but I imagine it's doable. I might attempt it later.

Edit: I'm starting to think this system isn't solvable in general...

3. It is solvable, I just cannot prove that in general the resulting expressions cannot get rid of trigonometric functions.

4. Originally Posted by elim
It is solvable, I just cannot prove that in general the resulting expressions cannot get rid of trigonometric functions.
Right, when I said it wasn't solvable, I meant expressing it without trig functions involved.

5. Originally Posted by chiph588@
You want to find $p,q\in\mathbb{R}$ such that $\sqrt[3]{a+bi} = p+qi \implies a+bi = (p+qi)^3 \implies p^3-3pq^2+(3p^2q-q^3)i = a+bi$

Therefore we have two equations and two unknowns:

$\begin{cases} p^3-3pq^2 = a\\3p^2q-q^3 = b \end{cases}$

Looks rather messy to solve, but I imagine it's doable. I might attempt it later.

Edit: I'm starting to think this system isn't solvable in general...
Ok, I'll show you in general you can't express $p$ and $q$ in terms of radicals.

From the two equations we can get $3p\sqrt{p^3-a}-\left(\frac{\sqrt{p^3-a}}{p}\right)^3 = b$ (solve for $q$ in equation 1 and substitute into equation 2).

Now square both sides:
$-\frac{a^3}{p^6}+\frac{3a^2}{p^3}-\frac{6a^2}{p^2}-9ap^2+12ap-3a+9p^5-6p^4+p^3=b^2$

Simplify:
$9p^{11}-6p^{10}+p^9-9ap^8+12ap^7-(3a+b^2)p^6-6a^2p^4+3a^2p^3-a^3=0$

Since there is no formula that can solve for a polynomial of degree greater than $4$ in terms of radicals, we see that this equation is unsolvable in general.

Applying this method to $q$ yields the same results.

6. Let $(a,b),(x,y) \in \mathbf{R}^2$ such that $a+ib=(x+iy)^3 = x^3-3xy^2+i(3x^2y-y^3)$
Then when $a=0$ we have $(x,y)=(0,-\sqrt[3]{b})$ as a solution.
If $a \neq 0$, then $x \neq 0$ and we have
$b^2 = (3x^2y-y^3)^2=\left(\frac{x^3-a}{3x}\right)\left(3x^2-\frac{x^3-a}{3x}\right)^2$ hence
$64x^9-48ax^6-(15a^2+27b^2)x^3-a^3=0$
Let $x^3=z+a/4, \quad p=-\frac{27}{64}(a^2+b^2),\quad q=-\frac{27a}{256}(a^2+b^2)$, then we have
$z^3+pz+q = 0$ and so z can be expressed in terms of radicals and so can x and y.

But the problem is that the radicals themselves(if using Cardano's method or known methods for me) are involving complex numbers and we get cyclic situations!

Seems we can express the problem as this:
Does real cubic equation $z^3+pz+q = 0$ solvable in real field?
Or how to disprove the equation is solvable in real field?

7. Originally Posted by elim
Let $(a,b),(x,y) \in \mathbf{R}^2$ such that $a+ib=(x+iy)^3 = x^3-3xy^2+i(3x^2y-y^3)$
Then when $a=0$ we have $(x,y)=(0,-\sqrt[3]{b})$ as a solution.
If $a \neq 0$, then $x \neq 0$ and we have
$b^2 = (3x^2y-y^3)^2=\left(\frac{x^3-a}{3x}\right)\left(3x^2-\frac{x^3-a}{3x}\right)^2$ hence
$64x^9-48ax^6-(15a^2+27b^2)x^3-a^3=0$
Let $x^3=z+a/4, \quad p=-\frac{27}{64}(a^2+b^2),\quad q=-\frac{27a}{256}(a^2+b^2)$, then we have
$z^3+pz+q = 0$ and so z can be expressed in terms of radicals and so can x and y.

But the problem is that the radicals themselves(if using Cardano's method or known methods for me) are involving complex numbers and we get cyclic situations!

Seems we can express the problem as this:
Does real cubic equation $z^3+pz+q = 0$ solvable in real field?
Or how to disprove the equation is solvable in real field?
Wow, didn't think of it this way!