# simplification when w is cube root of unity

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W^8 + w^2 has two solutions. One is -1 which I have achieved in many different ways. The other solution is 2. I cannot get it. Can anyone give me a clue? Thanks

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#2

Can you upload the original question?

Do you mean value of the expression or w^8+w^2=0 or ...

Do you mean value of the expression or w^8+w^2=0 or ...

Last edited by mqb2766; 6 months ago

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(Original post by

Can you upload the original question?

Do you mean value of the expression or w^8+w^2=0 or ...

**mqb2766**)Can you upload the original question?

Do you mean value of the expression or w^8+w^2=0 or ...

The question is: If w is the cube root of 1, find the possible values of each of the following: (the first 3 I managed to do) and then w^8 + w^10 There was no = 0 or = anything Here is the question it's form the book called Further Pure Mathematics by Brain and Mark Gaulter. I tried to get an image but it's turned out a bit too small I think

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#4

(Original post by

Sorry I made an error in typing

The question is: If w is the cube root of 1, find the possible values of each of the following: (the first 3 I managed to do) and then w^8 + w^10 There was no = 0 or = anything Here is the question it's form the book called Further Pure Mathematics by Brain and Mark Gaulter. I tried to get an image but it's turned out a bit too small I think

**maggiehodgson**)Sorry I made an error in typing

The question is: If w is the cube root of 1, find the possible values of each of the following: (the first 3 I managed to do) and then w^8 + w^10 There was no = 0 or = anything Here is the question it's form the book called Further Pure Mathematics by Brain and Mark Gaulter. I tried to get an image but it's turned out a bit too small I think

Note that w^3=w^6=w^9=1, for any root. So the expression can be simplified to...

Last edited by mqb2766; 6 months ago

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As you can see, for me all roads lead to minus 1. Can’t get it to 2. Any further clue?

Thanks

Thanks

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#6

It reduces to

w + w^2

The roots of unity are

1, -1/2+/-i*sqrt(3)/2

The two complex roots are conjgates.

...

You can also get it by factorizing

w^3 - 1 = 0

w + w^2

The roots of unity are

1, -1/2+/-i*sqrt(3)/2

The two complex roots are conjgates.

...

You can also get it by factorizing

w^3 - 1 = 0

Last edited by mqb2766; 6 months ago

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(Original post by

It reduces to

w + w^2

The roots of unity are

1, -1/2+/-i*sqrt(3)/2

The two complex roots are conjgates.

...

**mqb2766**)It reduces to

w + w^2

The roots of unity are

1, -1/2+/-i*sqrt(3)/2

The two complex roots are conjgates.

...

Thanks again, without your help I would have been struggling for even longer.

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#8

(Original post by

I got that it reduced to w + w^2. I have now found out a little extra (which is now obvious): if a cube root of 1 is 1 then w + w^2 2. Simple when you know how.

Thanks again, without your help I would have been struggling for even longer.

**maggiehodgson**)I got that it reduced to w + w^2. I have now found out a little extra (which is now obvious): if a cube root of 1 is 1 then w + w^2 2. Simple when you know how.

Thanks again, without your help I would have been struggling for even longer.

w^3 = 1

Is factorized as

(w - 1)(w^2 + w + 1) = 0

The quadratic factor corresponds to the two complex roots and both must satisfy

w^2 + w = -1

The linear factor is

w = 1

So

w^2 + w = 2

You don't need to solve for/know the comp!ex root values, but picturing them on the unit circle (modulus 1, arg 0, 120, 240) gives insight. In particular, why w^2 is the conjugate of w.

Last edited by mqb2766; 6 months ago

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(Original post by

Also solving

w^3 = 1

Is factorized as

(w - 1)(w^2 + w + 1) = 0

The quadratic factor corresponds to the two complex roots and both must satisfy

w^2 + w = -1

The linear factor is

w = 1

So

w^2 + w = 2

You don't need to solve for/know the comp!ex root values, but picturing them on the unit circle (modulus 1, arg 0, 120, 240) gives insight. In particular, why w^2 is the conjugate of w.

**mqb2766**)Also solving

w^3 = 1

Is factorized as

(w - 1)(w^2 + w + 1) = 0

The quadratic factor corresponds to the two complex roots and both must satisfy

w^2 + w = -1

The linear factor is

w = 1

So

w^2 + w = 2

You don't need to solve for/know the comp!ex root values, but picturing them on the unit circle (modulus 1, arg 0, 120, 240) gives insight. In particular, why w^2 is the conjugate of w.

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#10

(Original post by

I would like to ask another question on the same topic if you don’t mind. Question 7c from the image I attached gave just the answer to be w^2 which I got. Could a numerical answer have been one? I don’t see why not but it’s now bugging me that the book answer was not 1.

**maggiehodgson**)I would like to ask another question on the same topic if you don’t mind. Question 7c from the image I attached gave just the answer to be w^2 which I got. Could a numerical answer have been one? I don’t see why not but it’s now bugging me that the book answer was not 1.

(w+w^4)/(w^2+w^5)

If so, that is

1/w = w^2

So it could (numerically) be 1,-1/2+/-isqrt(3)/2 as if w is a root of unity, so is w^2.

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(Original post by

Is it

(w+w^4)/(w^2+w^5)

If so, that is

1/w = w^2

So it could (numerically) be 1,-1/2+/-isqrt(3)/2 as if w is a root of unity, so is w^2.

**mqb2766**)Is it

(w+w^4)/(w^2+w^5)

If so, that is

1/w = w^2

So it could (numerically) be 1,-1/2+/-isqrt(3)/2 as if w is a root of unity, so is w^2.

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#12

(Original post by

Thanks. I wonder why they used w^2 as an answer on this question and numbers on the others. None of the questions had the answers -1/2+/-isqrt(3)/2 but my book has explained where they come from and I understand it.

**maggiehodgson**)Thanks. I wonder why they used w^2 as an answer on this question and numbers on the others. None of the questions had the answers -1/2+/-isqrt(3)/2 but my book has explained where they come from and I understand it.

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