Complex help (1 Viewer)

[fullonoob]

i) find the five fifth roots of unity
ii) if w = cis 2pi/5, show that 1+w+w^2+w^3+w^4 = 0
iii) show that z1 = w +w^4 and z2 = w2+w3 are the roots of the equation z^2+z-1 = 0

Just help with ii) and iii) please
ii) is just sum of roots = 0 but i dont really get that still

 

[fullonoob]

the polynomial is w^5 - 1=0

then factorising that, you get (w-1)(1+w+w^2+w^3+w^4)=0

since w is not 1, but w is a root, the 1+w+w^2+w^3+w^4 must equal zero.

Yeah i kinda got that but how do you do iii)

 

[Trebla]

i) find the five fifth roots of unity
ii) if w = cis 2pi/5, show that 1+w+w^2+w^3+w^4 = 0
iii) show that z1 = w +w^4 and z2 = w2+w3 are the roots of the equation z^2+z-1 = 0

Just help with ii) and iii) please
ii) is just sum of roots = 0 but i dont really get that still

If roots are z₁ and z₂ then
Sum of roots:
z₁ + z₂ = w + w⁴ + w² + w³
= w + w² + w³ + w⁴
= - 1 (using the result in ii))

Product of roots:
z₁z₂ = (w + w⁴)(w² + w³)
= w³ + w⁴ + w⁶ + w⁷
= w³ + w⁴ + w(w⁵) + w²(w⁵)
= w + w² + w³ + w⁴
(since w⁵ = 1 as w satisfies a fifth root of unity)
= - 1 (using the result in ii))

We now have:
z₁ + z₂ = - 1
z₁z₂ = - 1

We can now deduce the quadratic equation using relationships between roots and coefficients:
z² + z - 1 = 0