another problem (complex #'s + more) (1 Viewer)

[spice girl]

1) Let A_1, A_2, ..., A_n represent the nth roots of unity w_1, w_2, ..., w_n. Suppose P represents z such that |z| = 1

(btw, w_1 is omega subscript 1, etc)

i) Prove w_1 + w_2 + ... + w_n = 0
ii) Show that |PA_i|^2 = (z-w_i)(z(bar) - w_i(bar)) (for all i = 1, 2, ..., n)
iii) Hence prove |PA_1|^2 + |PA_2|^2 + ... + |PA_n|^2 = 2n

 

[OLDMAN]

i) w's satisfy z^n-1=0 , therefore sum of w's =0.
ii) mod (PA_i) square = (z-w_i)(z-w_i)bar since IzI^2=z.zbar
=(z-w_i)(zbar-w_ibar)
iii) sum (z-w_i)(zbar-w_ibar) for i from 1 to n
= sum(z.zbar+w_i.w_ibar-w_iZbar-w_ibarZ) for i 1 to n
= 2n since mod z and mod w both =1, and both sum(w_i) and sum(w_ibar) for i 1 to n =0 note w_i and w_ibar are conjugate, roots of unity. Proven

 

[spice girl]

think that was a past HSC question...around question 6'ish...can't remember it's been so long...

 

[ND]

Originally posted by OLDMAN
i) w's satisfy z^n-1=0 , therefore sum of w's =0.

I don't really understand that, is it the same as this?:

let z^n - 1 = 0

since roots are w_1, w_2 etc. and sum of roots = 0/-1 :
w_1 + w_2 + ... + w_n = 0

 

[spice girl]

sum of roots to the polynomial z^n - 1 = 0 is 0 (because coefficient of z^(n-1) term is 0

 

[OLDMAN]

must admit I do skip steps. eg. sum(w_ibar)=0
which comes from sum(w_i)=0, therefore
sum(w_i)bar=0bar and thus the result.

 

[KeypadSDM]

i) z^n - 1 = 0

(z - 1)(z^(n-1) + z^(n-2) + ... + z^(n-n)) = 0

.: z^(n-1) + z^(n-2) + ... + z^1 + z^0 = 0

let F(z) = z^(n-1) + z^(n-2) + ... + z^1 + z^0
since, F(1) > 0
z =/= 1
.: all w's must be roots of this equation...

Can this somehow lead to the required answer?

actually i can see, the co efficient of the (n-1) term is 1
.: sum of w's is -1 ( i don't treat 1 as a w here)
.: sum of roots = 0