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December 29th, 2013, 03:39 AM   #1
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Cube roots of -1

Find all 3 roots of



In terms of only. Show all work.

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December 29th, 2013, 03:49 AM   #2
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Re: Cube roots of -1

Do you mean it like this?











Or do want to also to be in terms of i?
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December 29th, 2013, 04:01 AM   #3
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Re: Cube roots of -1

Quote:
Originally Posted by crom
Do you mean it like this?











Or do want to also to be in terms of i?
Thank you for your response.

I like what you did with



that is one solution in terms of

Can you transform the other 2 solutions so that they are both a SINGLE term of ?

Hint : You can go the long way or try to rethink the problem using rotations.

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December 29th, 2013, 05:39 AM   #4
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Re: Cube roots of -1

Dual to crom:



(Edited) Off the chain:



(Edited) Furthermore:

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December 29th, 2013, 07:22 AM   #5
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Re: Cube roots of -1

Quote:
Originally Posted by mathbalarka
Dual to crom:



(Edited) Off the chain:



(Edited) Furthermore:

Nice!

Ha - Ha , admittedly , I was looking for



which is equivalent to your 'off the chain' version but it's nice to see you both working the cheat kobayashi maru style , especially [color=#00FF00]mathbalarka [/color] who really went to town on it!!

BTW , since [color=#0000FF]Crom [/color] offered ... give me in terms of a single term of

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December 29th, 2013, 11:28 AM   #6
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Re: Cube roots of -1

Let's take then



( we also have as you said, but I will use this one )



then



Is this how you meant or?
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December 29th, 2013, 08:05 PM   #7
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Re: Cube roots of -1

Quote:
Originally Posted by crom
Let's take then



( we also have as you said, but I will use this one )



then



Is this how you meant or?
That's nice, I like it, but you are using multiple terms involving .


Let me give the final result and ask you (or anyone else interested) to provide the calculations, if you don't mind. Highlight below.

*^*[color=#FFFFFF]you can hide the answer by choosing the appropriate font color [/color][/color]

I have 'hidden' a word sentence and the latex fraction 1/3. To see it , quote my post , delete all the color tags only, then click Preview. If this is too much work then post it un-hidden , I don't mind.

The result I'm looking for is



Can you show the steps needed to get there? Note it is only one term using and two constants.

Sorry about the tiny exponent, I don't know how to enlarge the font inside latex.

In words, eye to the power of [(minus eye times the natural logarithm of three) divided by pi]

If anyone knows how to enlarge the font inside latex, please advise.

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December 30th, 2013, 02:57 AM   #8
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Re: Cube roots of -1

You could use \large, \Large, \huge or \Huge to enlarge LaTeX. Likewise \small or \tiny to make it smaller.
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December 30th, 2013, 04:06 AM   #9
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Re: Cube roots of -1

Quote:
Originally Posted by agentredlum
The result i'm looking for is ...
I'd represent it as



And transform everything to by the simple yet interesting identity



We can generalize this to any elementary number (i.e., an algebraic number with minimal polynomial having solvable Galois group), see Gauss sums and Kronecker-Weber theorem and the later for anybody who wants to dig deep in these things.
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December 30th, 2013, 04:09 AM   #10
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Re: Cube roots of -1

Quote:
Originally Posted by Hoempa
You could use \large, \Large, \huge or \Huge to enlarge LaTeX. Likewise \small or \tiny to make it smaller.
Thought I tried that, well, I'll try it again.

Quote:
Originally Posted by Hoempa
Bravo!

I actually did it a little differently,



Now, on RHS we want to change e to 3, get rid of , get rid of the negative and keep the 2 in the denominator for use as square root



And the result follows on the LHS.

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