Complex (1 Viewer)

[azureus88]

Given z^7 = 1 where z is not=1, deduce that z^3 + z^2 + z + 1 + 1/z + 1/z^2 + 1/z^3 = 0.

How would you set out this question without assuming z is the complex root with smallest argument (cause it only says z is not=1)? Im a bit rusty on this topic so any help would be nice?

thanks in advance

 

[Trebla]

Given z^7 = 1 where z is not=1, deduce that z^3 + z^2 + z + 1 + 1/z + 1/z^2 + 1/z^3 = 0.

How would you set out this question without assuming z is the complex root with smallest argument (cause it only says z is not=1)? Im a bit rusty on this topic so any help would be nice?

thanks in advance

z⁷ - 1 = 0
(z - 1)(z⁶ + z⁵ + z⁴ + z³ + z² + z + 1) = 0
Since z =/= 1, then
z⁶ + z⁵ + z⁴ + z³ + z² + z + 1 = 0
Divide both sides by z³ (and obviously z =/= 0) gives:
z³ + z² + z + 1 + 1/z + 1/z² + 1/z³ = 0

 

[azureus88]

k nice

 

[azureus88]

if they ask a question lik "show that if @ is one of the complex roots of..., then @^2, @^3, @^4... are the other complex roots"

can you just let @ be the root with smallest argument and go from there. or do you have to consider other cases?

 

[tommykins]

w (omega) is the root with the smallest argument.

 

[azureus88]

the question has alpha, not omega though

 

[BeDifferent]

Hey doing similar question but there is an extension to this question, let x=z+1/z
....
blah blah

i got something like x^3+x^2-2x-1=0

and then the last part is

hence deduce
cos pi/7*cos2pi/7*cos3pi/7=1/8

help please

 

[azureus88]

ok i'll start from the part where you got your roots of unity as 1, cis(2pi/7), cis(-2pi/7), cis(4pi/7), cis(-4pi/7), cis(6pi/7), cis(-6pi/7).

let w=cis(2pi/7)

(w + w^(-1))(w^2 + w^(-2))(w^3 + w^(-3)) = (2cos(2pi/7))(2cos(4pi/7))(2cos(6pi/7))

by expanding you get

1+(1+w+w^2+w^3+w^4+w^5+w^6)=(2cos(2pi/7))(-2cos(3pi/7))(-2cos(pi/7))

but (1+w+w^2+w^3+w^4+w^5+w^6)=0 by sum of roots of w^7 - 1=0

(cospi/7)(cos2pi/7)(cos3pi/7)=1/8

 

[Trebla]

if they ask a question lik "show that if @ is one of the complex roots of..., then @^2, @^3, @^4... are the other complex roots"

can you just let @ be the root with smallest argument and go from there. or do you have to consider other cases?

You have to take @ as a general root, not specifically the one with the smallest argument. Generally you wouldn't have to worry if it was the smallest argument or not, because there are ways without considering the argument at all.

 

[BeDifferent]

Bump the question =]