Real algebra expression for the Re and Im of a cubic root

elim · March 9th, 2010, 06:14 AM

Let $a, b \in \mathbf{R}$ , one can easily check that
If $x,y \in \mathbf{R}$ and $(x+iy)^2= a+ib$ ,
then $x+iy= \pm \frac{1}{\sqrt{2}} \left( \sqrt{a+\sqrt{a^2+b^2}} + i \text{sgn}(b+) \sqrt{-a+\sqrt{a^2+b^2}}\right)$
Where $\text{sgn}(b+)= 1$ when $b \geq 0$ , otherwise $\text{sgn}(b+)= -1$

We see that the real and imaginary parts of a square root of a complex number can always given by real algebra expressions without trigonometric functions.

Now the question: Is it possible to express the real and imaginary parts of a cubic root of a complex number the same manner?

My guess is no. But anyone can give a solid argument on this?

cknapp · March 10th, 2010, 10:22 PM

I would say yes, because the cubic is solvable... but I'm too lazy right now to figure out what the messy closed-form solution is.

elim · March 17th, 2010, 01:34 PM

I can only obtain the closed form with trigonometric function involved

g_litched · March 23rd, 2010, 11:05 AM

Cubes are all square and therefore are not as strong as using odd numbers.
so raising numbers and keeping everything square keeps a balance because they are divisble by 2.

elim · March 26th, 2010, 12:47 PM

I don't understand g_litched's point

cknapp · April 3rd, 2010, 02:20 PM

Neither can I, frankly. It sounds like nonsense... as with the rest of his posts.

Anyway, I'm not sure if my justification above for why it should be will really carry through; it was more of an educated guess.

Would I be able to see one of your attempts (that gives the solution with trig functions), and see if I can't try to do something with that?

elim · April 4th, 2010, 03:20 PM

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?

March 9th, 2010, 06:14 AM	#1
elim Senior Member Joined: Dec 2006 Posts: 167 Thanks: 3	Real algebra expression for the Re and Im of a cubic root Let $a, b \in \mathbf{R}$ , one can easily check that If $x,y \in \mathbf{R}$ and $(x+iy)^2= a+ib$ , then $x+iy= \pm \frac{1}{\sqrt{2}} \left( \sqrt{a+\sqrt{a^2+b^2}} + i \text{sgn}(b+) \sqrt{-a+\sqrt{a^2+b^2}}\right)$ Where $\text{sgn}(b+)= 1$ when $b \geq 0$ , otherwise $\text{sgn}(b+)= -1$ We see that the real and imaginary parts of a square root of a complex number can always given by real algebra expressions without trigonometric functions. Now the question: Is it possible to express the real and imaginary parts of a cubic root of a complex number the same manner? My guess is no. But anyone can give a solid argument on this?
$elim is offline$

March 10th, 2010, 10:22 PM	#2
cknapp Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3	Re: Real algebra expression for the Re and Im of a cubic root I would say yes, because the cubic is solvable... but I'm too lazy right now to figure out what the messy closed-form solution is.
$cknapp is offline$

March 17th, 2010, 01:34 PM	#3
elim Senior Member Joined: Dec 2006 Posts: 167 Thanks: 3	Re: Real algebra expression for the Re and Im of a cubic root I can only obtain the closed form with trigonometric function involved
$elim is offline$

March 23rd, 2010, 11:05 AM	#4
g_litched Newbie Joined: Mar 2010 Posts: 9 Thanks: 0	Re: Real algebra expression for the Re and Im of a cubic root Cubes are all square and therefore are not as strong as using odd numbers. so raising numbers and keeping everything square keeps a balance because they are divisble by 2.
$g_litched is offline$

March 26th, 2010, 12:47 PM	#5
elim Senior Member Joined: Dec 2006 Posts: 167 Thanks: 3	Re: Real algebra expression for the Re and Im of a cubic root I don't understand g_litched's point
$elim is offline$

April 3rd, 2010, 02:20 PM	#6
cknapp Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3	Re: Real algebra expression for the Re and Im of a cubic root Neither can I, frankly. It sounds like nonsense... as with the rest of his posts. Anyway, I'm not sure if my justification above for why it should be will really carry through; it was more of an educated guess. Would I be able to see one of your attempts (that gives the solution with trig functions), and see if I can't try to do something with that?
$cknapp is offline$

April 4th, 2010, 03:20 PM	#7
elim Senior Member Joined: Dec 2006 Posts: 167 Thanks: 3	Re: Real algebra expression for the Re and Im of a cubic roo 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?
$elim is offline$

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