# Thread: cube roots of unity

1. ## cube roots of unity

Hi folks,

there is a very well worn method of finding the cube roots of unity, but there are always more than one way of doing things, so I tried a few, hoping that all roads lead to the mathematical rome.

$z^3 = 1$

$(x + iy)^3 = 1$

$x^3 - 3xy^{2} + 3x^{2}yi -iy^{3} = 1$

$(x^3 - 3xy{^2}) + i(3x{^2}y - y^3) = 1$

so

$x^3 - 3xy^{2} = 1$ ...... (1)

$3x^{2}y - y^3 = 0$ .......(2)

from (2)

$y ( y^2 - 3x^2) = 0$

$y^2 = 3x^2$

from (1)

$x ( x^2 - 3y^2) = 1$

$x = 1$ or $x^2 = 3y^2 + 1$

$x^2 = 9x^2 + 1$

$-8x^2 = 1$

this is not going well, but I can't see the flaw in it. Where am I going wrong?

2. ## Re: cube roots of unity

Originally Posted by s_ingram
from (1)

$x ( x^2 - 3y^2) = 1$

$x = 1$ or $x^2 = 3y^2 + 1$

$x^2 = 9x^2 + 1$

$-8x^2 = 1$

this is not going well, but I can't see the flaw in it. Where am I going wrong?
$x(x^2-3y^2) = 1$

From (2), you found $y=0$ or $y^2=3x^2$.

If $y=0$, then you have $x^3=1$, which means $x=1$ since $x\in \mathbb{R}$.
If $y^2=3x^2$, then you have:
$x(x^2-9x^2) = 1$
$-8x^3 = 1$
$x^3 = -\dfrac{1}{8}$
$x = -\dfrac{1}{2}$

Plugging back in, we have $y^2 = 3x^2 = \dfrac{3}{4}$ so $y = \pm \dfrac{\sqrt{3}}{2}$

The problem with your method is that $ab=1$ does NOT imply $a=1$ or $b=1$. Example: $2\cdot \dfrac{1}{2} = 1$ and neither factor is 1. You were confusing this with $ab=0$. This implies $a=0$ or $b=0$.

3. ## Re: cube roots of unity

Originally Posted by s_ingram
Hi folks,

there is a very well worn method of finding the cube roots of unity, but there are always more than one way of doing things, so I tried a few, hoping that all roads lead to the mathematical rome.

$z^3 = 1$

$(x + iy)^3 = 1$

$x^3 - 3xy^{2} + 3x^{2}yi -iy^{3} = 1$

$(x^3 - 3xy{^2}) + i(3x{^2}y - y^3) = 1$

so

$x^3 - 3xy^{2} = 1$ ...... (1)

$3x^{2}y - y^3 = 0$ .......(2)

from (2)

$y ( y^2 - 3x^2) = 0$

$y^2 = 3x^2$
Actually, either $y^2= 3x^2$ or y= 0. y= 0 reduces (1) to $x^2= 1$ which gives x= 1.

6. ## Re: cube roots of unity

yes, I assumed it was just a typo. Also, after all the help I got, it seemed kind of rude to be so picky. But you guys all all mathematicians, precision really counts. In future I will be more rigerous!

7. ## Re: cube roots of unity

No, not rude at all, just a helpful correction.

(Would it be rude of me to point out that "rigorous" is not spelled with a "e"?)

8. ## Re: cube roots of unity

Ah... you see! You got me again!