complex root of unity (1 Viewer)

[pLuvia]

Prove the follwoing, if w is a complex cube root of unity

(a+b)(aw+bw²)(aw²+bw) = a³+b³

(a+wb+w²c)(a+w²b+wc) = a²+b²+c²-ab-ac-bc

 

[VivianHsu]

(a+b)(aw+bw2)(aw2+bw) = a3+b3

a3+b3 = (a+b)(a2-ab+b2)
so need to prove (aw+bw2)(aw2+bw) = (a2-ab+b2)
LHS = a2w3 + abw4 + abw2 + b2w3
= a2w3 + abw(w3) + abw2 + b2w3 then on using w3=1,
= a2 + abw + abw2 + b2
= a2 + ab(w + w2) + b2
but w+w2+w3=0 (sum of roots = 0) so w+w2=-1



(a+wb+w^2c)(a+w^2b+wc)
=a^2+abw^2+acw+abw+b^2+bcw^2+acw^2+bcw+c^2
=a^2+b^2+c^2+ab(w+w^2)+ac(w+w2)+bc(w+w^2)
=a^2+b^2+c^2-ab-ac-bc

 

[pLuvia]

but w+w2+w3=0 (sum of roots = 0) so w+w2=-1

I don't understand this bit

 

[香港!]

^
w^3=1


123156165

 

[pLuvia]

^
w^3=1


123156165

what are you trying to say?

 

[VivianHsu]

I don't understand this bit

sum of all complex roots of unity (and of anything actually) = 0
so sum of all complex cube roots of unity = 0
these roots are
1, cos2pi/3+isin2pi/3, cos2pi/3-isin2pi/3

in the special case of roots of unity (1), all of the roots can be expressed as powers of one of the roots with an imaginary part.
for example if w = cos2pi/3+isin2pi/3
then w^2 = cos2pi/3-isin2pi/3
and w^3 = 1

if w = cos2pi/3-isin2pi/3 then w^2 = cos2pi/3+isin2pi/3 (check it if u want)

in this q, w has to be a root with an imaginary part... probably the q isnt very clear

 

[pLuvia]

How about these questions

Solve:

(z+1)⁴ + 16 = 0
(1-i)z - 2i = 2
(z-1)³ + 8(z+i)³=0

Prove if w³=1

(2+5w+2w²)⁶ = 729

 

[VivianHsu]

(z+1)^4 + 16 = 0
(z+1)^4 = -16
(z+1)^4 = 16 cis(pi+2npi), n integer
(z+1) = 16^(1/4) cis[1/4(pi+2npi)]
= 2 cis[1/4(pi+2npi)]
then put n=0,1,2,3 to get 4 roots in total


(1-i)z - 2i = 2
(1-i)z = 2(1+i)
z=2(1+i)/(1-i)
z=2(1+i)/(1-i) * (1+i)/(1+i)


(z-1)^3 + 8(z+i)^3=0
[(z-1)/(z+i)]^3 = -8 = 8cis(pi/2+2npi)
(z-1)/(z+i) = a, b, c (3 things)
say for a, (z-1)/(z+i) = a
(z-1) = a(z+i) => z(1-a)=ai+1 etc


(2+5w+2w2)6 = 729
2+5w+2w2
= 2w3+5w+2w2
= 2w3 + 2w + 2w2 + 3w
= 2(w3+w+w2)+3w
= 0+3w (sum of roots of unity = 0)
then (2+5w+2w2)6 = (3w)6 = 729w6 = 729(w3)2 = 729(1)^2

 

[boredguy]

i see you go to north shore

 

[Riviet]

i see you go to north shore

Haha, i deduce that you must too! Probably some homework questions lol.

 

[pLuvia]

i see you go to north shore

hehe yeh which one do you go to?

 

[boredguy]

the chatswood branch

 

[pLuvia]

Really? lol, Saturday morning? You have that chinese teacher? the one that always says "is it" at the end of every sentence?

 

[boredguy]

yea i have him lol
u go there too?

 

[pLuvia]

yup

I don't like him too much

 

[boredguy]

sat morning or afternoon?

 

[pLuvia]

morning 11am

 

[Riviet]

Hehe it's a small world we live in... by the way kadlil, it IS the RYDA thing that i'm going to this friday, have you been to it? (what's ryda stand for anyway?)

 

[Shrutzzz]

Help!!!! : Complex Root of unity question

Could anyone please help me solve this Cambride Book qs Exercise 2.4) question 4

Use DeMoivre's theorem to solve z^5 =1 by grouping the roots in complex conjugate pairs ,show that

z^5 +1 = (z+1)(z^2-2zcos pi/5 +1 ) (z^2-2zcos 3pi/5 +1 )


Thank youuuuuuuuuuuu

Shrutzzzz

 

[Sober]

Could anyone please help me solve this Cambride Book qs Exercise 2.4) question 4

Use DeMoivre's theorem to solve z^5 =1 by grouping the roots in complex conjugate pairs ,show that

z^5 +1 = (z+1)(z^2-2zcos pi/5 +1 ) (z^2-2zcos 3pi/5 +1 )


Thank youuuuuuuuuuuu

Shrutzzzz

You wrote the wrong question, surely it must be z^5 = -1 ?

z = (-1)^(1/5) = cis( (1 + 2k) pi / 5 )

The roots are (underline denotes complex conjugate):

z0 = cis(pi / 5)
z1 = cis(3 pi / 5)
z2 = cis(pi) = -1
z3 = cis(7 pi / 5) = cis(-3pi / 5) = z1
z4 = cis(9 pi / 5) = cis(-pi / 5) = z0

0 = (z-z0)(z-z1)(z-z2)(z-z3)(z-z4)
= (z + 1)(z - z0)(z - z0)(z - z1)(z - z1)
= (z + 1)(z^2 + 1 + 2zcos(pi / 5) )( z^2 + 1 + 2zcos(3 pi / 5) )