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 December 30th, 2013, 06:31 AM #21 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Cube roots of -1 I don't want to post anything there because I would prefer to keep that post clean for any wandering transcendental number theorist that may happen to pass by. For your first bullet (by first bullet, I am referring to the top bullet), you are asking about the kobayashi maru style cheats (which I like all of them BTW) presented by [color=#0000FF]crom[/color], [color=#00FF00]mathbalarka[/color] and [color=#00FF00]Hoempa[/color]? If yes, then I think there should be many (perhaps infinite) ways to represent root 3. I don't think I understand the second bullet. Is $\ \ln(i) \$ an algebraic number? Too tired to look it up... If not, then $\sqrt{3} \= \ \Large i^{\frac{\ln3}{\ln i}$ But like I said, I'm not sure about what you are asking, so bear with me.
 December 30th, 2013, 08:17 AM #22 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Cube roots of -1 $\log$$i$$= \frac{\pi}{2}$ Also, note that $^{\log 3}$is also transcendental by Lindemann-Weierstrass theorem. But that does not prove whether or not $^{\log(3)/\log(i)}$ is algebraic or not. $\frac{\log 3}{\log i}= \frac{\log 3}{\pi/2} = \log \left(3^{2/\pi}\right)$ And $3^{2/\pi}$ is transcendental by Gelfond–Schneider, unfortunately, so I cannot prove or disprove whether your expression holds or not. PS : That should have been $i^{\frac{\log 3}{\log ( -1)}}$ but similar consideration holds and we have nothing much. PPS : You can post this on that topic, as I believe it is non trivial enough to get considerations from those transcendental passer by s... Oops, I mean transcendental number theorist passer-by s...
December 30th, 2013, 10:34 PM   #23
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$\text{As }-1=i^{\small2}=i^{\small6}=i^{\small10},\text{ its cube roots are }i^{\small2/3},\ i^{\small6/3}\text{ and }i^{\small10/3}.$

When using latex, type \ln rather than just ln for the ln function.

Quote:
 Originally Posted by mathbalarka $\log$$i$$= \frac{\pi}{2}$
Incorrect.

December 30th, 2013, 11:37 PM   #24
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Re: Cube roots of -1

Quote:
 Originally Posted by skipjack Incorrect.
Right, $\frac{\pi}{2} \cdot \box{i}$. I am not sure whether this would effect anything (most probably would), I have to take a look at it seriously later.

December 31st, 2013, 02:57 AM   #25
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Re:

Quote:
 Originally Posted by skipjack $\text{As }-1=i^{\small2}=i^{\small6}=i^{\small10},\text{ its cube roots are }i^{\small2/3},\ i^{\small6/3}\text{ and }i^{\small10/3}.$
As usual [color=#BF0000]skipjack[/color] has posted the best answer so far. I did it a little bit differently, first, as a matter of personal preference,

$i^{ \small10/3} \= \ i^{ \small3} \cdot i^{ \small1/3} \ = \ i^{ \small2} \cdot i \cdot i^{ \small1/3} \ = \ -i^{ \small3/3} \cdot i^{ \small1/3} \ = \ -i^{ \small4/3}$

And that's what I used.

Second, if you know one root you can get the others by using symmetry and rotations.

Multiplication by $\ i \$ rotates a point counterclockwise in the complex plane by 90°.

Multiplication by $\ \sqrt{i} \$ rotates a point counterclockwise in the complex plane by 45°.

...etc.

In general, if you want to rotate counterclockwise by N°, multiply by

$i^{N/90}$

So, I know one root of $\ x^3 \= \ -1 \ \$ is x = -1.

I know the roots are symmetrically located at intervals of 120° = 360°/3.

I can use $\ i^{ \small120/90} \= \ i^{ \small4/3} \$ as a rotation machine.

To rotate x = -1, multiply it like so,

$-1 \cdot i^{ \small4/3} \= \ -i^{4/3}$

There's another root, now do it again,

$-i^{ \small4/3} \cdot i^{ \small4/3} \= \ -i^{ \small8/3} \ = \ -i^{ \small2} \cdot i^{ \small2/3} \ = \ i^{ \small2/3}$

And there's the other root.

Anyway, hope you see how to play around with multiplications additions and rotations.

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