Sy123
This too shall pass
- Joined
- Nov 6, 2011
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- HSC
- 2013
a) Find, in modulus-argument form, the roots of the equation z^(2n+1) =1.
b) Hence factorise z^2n + z^(2n-1) + ... + z^2 + z +1 into quadratic factors with real coefficients.
c) Deduce that 2^n * sin(pi/(2n+1)) * sin(2pi/(2n+1)) * sin(3pi/(2n+1)) ... sin(n*pi/(2n+1)) = sqrt(2n+1)
let P(x) = x^2-px +q =0Nice work.
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Nice worklet P(x) = x^2-px +q =0
if a and b are roots, therefore P(a) = P(b)=0
P(a) = a^2 -pa+q=0, times by a^n
P(b) = b^2 -pb +q=0, times by b^n
therefore:
a^(n+2) -pa^(n+1) +qa^n =0
b^(n+2) -pb^(n+1) +qb^n =0
solving simultaneously by ADDITION:
a^(n+2) +b^(n+2) -p(a^(n+1) +b^(n+1)) +q(a^n +b^n) =0
S_n =a^n +b^n
S_n+1 =a^(n+1) +b^(n+1)
S_n+2 =a^(n+2) +b^(n+2)
therefore S_n+2 -pS_n+1 + qS_n =0
ALTERNATIVELY,
since a and b are roots of P(x) = 0, then P(a)=P(b)=0
P(a) = a^2 -pa +q =0............[1]
P(b) = b^2 -pb +q =0.............[2]
S_n+2 -pS_n+1 + qS_n =0
a^(n+2) +b^(n+2) -p[a^(n+1) +b^(n+1)] +q(a^n +b^n) =0
a^(n+2) -pa^(n+1) +qa^n +b^(n+2) -pb^(n+1) +qb^n =0
a^n[a^2 -pa +q] + b^n[b^2 -pb +q] =0
by applying [1] and [2]
0+0=0
LHS=RHS therefore, S_n+2 -pS_n+1 + qS_n =0
Wow haha, not the solution I had in mind it can be done in 3 lines with cosine rule of non-right angled triangles. Either way good insight.
dam nvm forgot the indentity zzconj =|z|^2........Nice work
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Nope, everything is there. (think vector addition/subtraction or in twinklegal's case identities)Is there any more information?
haha, it was the first thing that came into my mind when I saw all those moduliWow haha, not the solution I had in mind it can be done in 3 lines with cosine rule of non-right angled triangles. Either way good insight.
(I would love another one of those really hard questions even if I cant solve it I want to have a go)
Yeah Realise asked something similar before and I answered it. I will post a question soon if I can make/find a good one.haha, it was the first thing that came into my mind when I saw all those moduli
I think a similar question has been asked before, not sure lol
not sure if this is right lol. a, b and c are meant to be integers as well right?Here is an easy (?) result I just proved.
Yes, well divisibility implies integers. But you got it right anyway.not sure if this is exactly right lol. Are a, b and c are meant to be integers as well? Or just rational/real?
i) sub z and by DE MOIVRE'S...Yes, well divisibility implies integers. But you got it right anyway.
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When I say cos k theta, I mean express cos^4 theta as cos theta, cos 2theta, etc
(Still waiting for a good question)
Correct, good job.i) sub z and by DE MOIVRE'S...
LHS =z^n + z^-n
= cosn(theta) + isinn(theta) + cosn(theta) -isinn(theta) = 2cos ntheta as required
ii) (z+1/z)^4 = z^4 +1/z^4 + 6(z^2+1/z^2) +6
noting z^n +1/z^n = 2cosntheta
threfore, 16cos^4theta = 2cos4theta + 8cos2theta +6
cos^4theta = 1/8 cos4theta + 0.5 cos2theta + 3/8