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March 9th, 2010, 06:14 AM   #1
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Real algebra expression for the Re and Im of a cubic root

Let , one can easily check that
If and ,
then
Where when , otherwise

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?
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March 10th, 2010, 10:22 PM   #2
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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.
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March 17th, 2010, 01:34 PM   #3
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Re: Real algebra expression for the Re and Im of a cubic root

I can only obtain the closed form with trigonometric function involved
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March 23rd, 2010, 11:05 AM   #4
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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.
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March 26th, 2010, 12:47 PM   #5
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Re: Real algebra expression for the Re and Im of a cubic root

I don't understand g_litched's point
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April 3rd, 2010, 02:20 PM   #6
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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?
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April 4th, 2010, 03:20 PM   #7
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Re: Real algebra expression for the Re and Im of a cubic roo

Let such that
Then when we have as a solution.
If , then and we have
hence

Let , then we have
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 solvable in real field?
Or how to disprove the equation is solvable in real field?
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