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 December 29th, 2013, 03:39 AM #1 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 Cube roots of -1 Find all 3 roots of $x^3 \= \ -1$ In terms of $\ i \$ only. Show all work.
 December 29th, 2013, 03:49 AM #2 Senior Member   Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11 Re: Cube roots of -1 Do you mean it like this? $x^3=-1$ $x^3+1=0$ $(x+1)(x^2-x+1)=0$ $x_1=-1=i^2$ $x_{2/3}=\frac{1 \pm i \sqrt3}{2}=\frac{-i^2 \pm i \sqrt3}{2}$ Or do want to $\sqrt3$ also to be in terms of i?
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? $x^3=-1$ $x^3+1=0$ $(x+1)(x^2-x+1)=0$ $x_1=-1=i^2$ $x_{2/3}=\frac{1 \pm i \sqrt3}{2}=\frac{-i^2 \pm i \sqrt3}{2}$ Or do want to $\sqrt3$ also to be in terms of i?

I like what you did with

$i^2$

that is one solution in terms of $\ i \$

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

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

 December 29th, 2013, 05:39 AM #4 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Cube roots of -1 Dual to crom: $\frac{ i \pm i\sqrt{i\cdot i+i\cdot i+i\cdot i}}{i+i}$ (Edited) Off the chain: $i^{\pm \frac{i + i}{i + i + i}$ (Edited) Furthermore: $\cos \left ( \frac{\log(i\cdot i) }{i+i+i}\right ) \pm \sin \left ( \frac{ \log(i\cdot i) }{i+i+i}\right )$
December 29th, 2013, 07:22 AM   #5
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Re: Cube roots of -1

Quote:
 Originally Posted by mathbalarka Dual to crom: $\frac{ i \pm i\sqrt{i\cdot i+i\cdot i+i\cdot i}}{i+i}$ (Edited) Off the chain: $i^{\pm \frac{i + i}{i + i + i}$ (Edited) Furthermore: $\cos \left ( \frac{\log(i\cdot i) }{i+i+i}\right ) \pm \sin \left ( \frac{ \log(i\cdot i) }{i+i+i}\right )$
Nice!

Ha - Ha , admittedly , I was looking for

$\{i^{\tiny{\frac{2}{3}}} \ , \ -i^{ \tiny{\frac{4}{3}}} \ , \ i^{ \tiny{\frac{6}{3}} \}$

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 $\ \sqrt{3} \$ in terms of a single term of $\ i \$

 December 29th, 2013, 11:28 AM #6 Senior Member   Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11 Re: Cube roots of -1 Let's take then $x^3=-1=i^2$ $x=i^{\frac{2}{3}}$ ( we also have $x= -i^{\frac{4}{3}}$ as you said, but I will use this one ) $\frac{1 +i \sqrt3}{2}=i^{\frac{2}{3}}$ then $\sqrt3=\frac{i^{\frac{2}{3}}+i^{\frac{2}{3}}+i^2}{ i}$ Is this how you meant or?
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 $x^3=-1=i^2$ $x=i^{\frac{2}{3}}$ ( we also have $x= -i^{\frac{4}{3}}$ as you said, but I will use this one ) $\frac{1 +i \sqrt3}{2}=i^{\frac{2}{3}}$ then $\sqrt3=\frac{i^{\frac{2}{3}}+i^{\frac{2}{3}}+i^2}{ i}$ Is this how you meant or?
That's nice, I like it, but you are using multiple terms involving $\ i \$ .

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 $[color=#FFFFFF] \frac{1}{3}$ [/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

$i^{ \tiny{-\frac{i\ln3}{ \pi}}} \= \ \sqrt{3}$

Can you show the steps needed to get there? Note it is only one term using $\ i \$ 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.

 December 30th, 2013, 02:57 AM #8 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Cube roots of -1 You could use \large, \Large, \huge or \Huge to enlarge LaTeX. Likewise \small or \tiny to make it smaller. $\sqrt{3}= e^{\large \frac{\ln(3)}{2}} = (i^{\large \frac{-2i}{\pi}})^{\large \frac{\ln(3)}{2}} = i^{\large \frac{-i \ln(3)}{\pi}}$
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

$\sqrt{3}= -ie^{i \pi/3} + ie^{-i\pi/3}$

And transform everything to $^{i}$ by the simple yet interesting identity

$e^\pi= i^{-2i}$

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.

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 $\sqrt{3}= e^{\large \frac{\ln(3)}{2}} = (i^{\large \frac{-2i}{\pi}})^{\large \frac{\ln(3)}{2}} = i^{\large \frac{-i \ln(3)}{\pi}}$
Bravo!

I actually did it a little differently,

$\Large i ^{i} \= \ \large e^{ \tiny{-} \large \frac{ \pi}{2}}$

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

$$$\Large i^{ \Large i}$$^{ \Large { \large \frac{ \tiny{-} ln 3}{ \pi}} \= \ $$\Large {e} ^{ \tiny{-} \frac{ \pi}{2}}$$^{ \tiny{-} \frac{ ln 3}{ \pi} }$

And the result follows on the LHS.

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