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FP3 nth roots of unity question

if w=(2pi/5)i simplify (1-w)(1-w2)(1-w3)(1-w4)

the answer is 5

I tried expanding and figured that was a waste of time

I have a feeling the answer is obvious please help
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Blazy
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Original post by physics4ever
if w=(2pi/5)i simplify (1-w)(1-w2)(1-w3)(1-w4)

the answer is 5

I tried expanding and figured that was a waste of time

I have a feeling the answer is obvious please help


There's probably better ways of doing it, but here we go:

Note that w5=1 w^5 = 1 and that 1−w5=(1−w)(1+w+w2+w3+w4) 1-w^5 = (1-w)(1+w+w^2+w^3+w^4) and so 1+w+w2+w3+w4=0 1+w+w^2+w^3+w^4 = 0 (since w≠1 w \neq 1 )

Expand (1−w)(1−w4) (1-w)(1-w^4) and (1−w2)(1−w3) (1-w^2)(1-w^3) separately (and noting my first point, simplify both expressions), and then multiply them together. You will end up with w6 w^6 and w7 w^7 terms but they can be simplified as well. Group bits together that look useful, e.g. 1+w+w2+w3+w4=0 1+w+w^2+w^3+w^4 = 0 and see where you end up :smile:
(edited 10 years ago)

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