Question

Cubic roots of 1 (complex numbers)

Hi,

How should I go on about solving the cubic roots of 1 or (omega) on my TI 89T.

I looked it up and I couldn't find anything about it, I dunno if I'm using the correct terms or not. (I don't study in English)

Anyways, here goes:
(the sum) => w+w^2+1=0
(multiply) => w*w^2*1 = 1

How do I go on about solving this:
(2+5w+2w^2)^2

By hand the correct answer (I hope lol) should be 9w^2

Thanks!

Answer 1

Are you trying to expand it, factor it, or solving for w?

Answer 2

I would suggest first expanding it, and then substitute u = w^2 to factor it.

Answer 3

Hm, I dunno.
I defined \[\omega = -\frac{ 1 }{ 2 }+\frac{ \sqrt{3} }{ 2 }i\]

then I calculated it and the answer is correct, it was:

\[(2+5\omega+2\omega^2)^2\] (sub w for what we defined) and the answer (on calc screen) is: \[-\frac{ 9 }{ 2 }-\frac{ 9\sqrt{3} }{ 2 }i\]

divide by w^2 and you get 9.

So it does calculate them when I define it into it, but it would be a lot better if I could make it display it symbolically instead of sub'ing.

Anyway to do that?

Answer 4

It's not clear (to me) what you are trying to answer.

if the question is
How should I go on about solving the cubic roots of 1

use the idea that
1 = exp( n 2 pi i) for n=1,2,3

with n=1, write 1 as exp( i 2 pi)
take the cube root: exp(i ⅔ pi)
convert exp( i A) to rectangular coords using cos(A) + i sin(A)

do the same thing for n=2 and n=3 (n=4 will repeat these 3 solutions)

Answer 5

Finding the cube roots of 1 isn't what we're looking for here.

We know that cube root of 1 has 3 roots:

\[1, -\frac{ 1 }{ 2 }+\frac{ \sqrt{3} }{ 2 }i,-\frac{ 1 }{ 2 }-\frac{ \sqrt{3} }{ 2 }i\]

so we replace the two imaginary roots with a w, and 1.
1, w, w^2 (order)

sum of all = 0
multiply of all = 0

and the question comes in the form of w equation, like the one I posted.
I don't know how to explain it good enough tbh

Answer 6

ok, but I assume you mean
multiply of all = 1
?

Answer 7

yeah, sorry, my bad.

Answer 8

so, you are asking,
given the two equations
1+w+w^2 = 0
1*w*w^2 = 1

how do we find all 3 unique solutions ?

Answer 9

Given those two equations, evaluate (2+5w+2w^2)^2

Answer 10

The correct answer should be 9w^2
my original question was to how to evaluate this using TI 89 lol

Answer 11

If you mean symbolically, no idea
if you mean numerically, then the manual will help
http://brownmath.com/ti83/complx89.htm

Answer 12

----------------------------------
w+w^2+1=0
w*w^2*1=0
----------------------------------
I'll put up an example for clarification.

If 1, w and w^2 are the cubic roots of 1, prove that:

\[(1+\omega^2)^8=\omega^2\]

We know that 1+w^2 = -w
substitute in the equation and evaluate (-w)^8
which is: w^8 which equals to w^2

I would like the TI 89 to do that, if possible.