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# Check my spurious algebra! watch

1. I'm working on this P6 question, roots of unity isn't very common on my syllabus so excuse me if I seem stupid on the topic.

Express z^5 -1 as the product of two factors.
z^5 - 1 = (z-1)(z^4 + z^3 + z^2 + z + 1)

given that a = e^(2/5)pi*i
Express the 5th roots of unity in terms of a
5th roots of unity are a cyclic group with generator a so the 5th roots of unity are
a, a^2, a^3, a^4, a^5

Show that the sum of the roots of unity is zero
Geometric progression, first term = a, r = a, n = 5

S(5) = a(1-a^5)/1-a

but a^5 = 1 so s(5) = 0

Find the exact value of (1 + a)(1 + a^2)(1 + a^3)(1 + a^4)
= (1 + a + a^2 + a^3)(1 + a^3 + a^4 + a^7)
= (-a^4)(1 + a^3 + a^4 + a^7)
= -(a^4 + a^7 + a^8 + a^11)

but
a^7 = a^2
a^8 = a^3
a^11 = a

so the expansion = -(a^4 + a^2 + a^3 + a)

= -(-a^5) = a^5 = e^2pi*i = 1

Is this correct? Very much appreciated if you can spot any holes or confirm it's all okay.
2. With the second part I'd probably have said the roots were a^0, a, a^2, a^3, a^4, but yours is also right as a^5 = 1 = a^0. It all looks ok though
3. (Original post by Bezza)
With the second part I'd probably have said the roots were a^0, a, a^2, a^3, a^4, but yours is also right as a^5 = 1 = a^0. It all looks ok though
For the last bit you could just notice that since z^4 + z^3 + z^2 + z + 1 = 0 has roots a, a^2, a^3, a^4 it can be written as (a - z)(a^2 - z)(a^3 - z)(a^4 - z) so the expression you want is the value when z = -1, which is 1 - 1 + 1 - 1 + 1 = 1

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