The Student Room Group

Some FP2 Help Please [MEI]



I completely do not understand part i) and ii). How did they draw that conclusion? Im completely lost.
Reply 1
Part i) Have you written out a few terms of C+iS?

Part ii) Double angle formulae.
Reply 2
for part i I have, 1 + (n1)e^jtheta + (n2)e^2jtheta etc
Reply 3
and e2iθ=(eiθ)2e^{2 i \theta}=(e^{i \theta})^2 so you have

1+(n1)eiθ+(n2)(eiθ)2+...1+\binom{n}{1}e^{i\theta}+\binom{n}{2}(e^{i \theta})^2+...
Reply 4
Original post by BabyMaths
and e2iθ=(eiθ)2e^{2 i \theta}=(e^{i \theta})^2 so you have

1+(n1)eiθ+(n2)(eiθ)2+...1+\binom{n}{1}e^{i\theta}+\binom{n}{2}(e^{i \theta})^2+...


ahh, im not sure how that becomes (i+e^jtheta)^n though =/
Reply 5
Original post by TimetoSucceed
ahh, im not sure how that becomes (1+e^jtheta)^n though =/


Expand (1+eiθ)n(1+e^{i \theta})^n then.

It should become clear.
Reply 6
If you expand that wouldn't you just get 1+e^jntheta, im finding this very difficult
Reply 7
maybe the (n 1), (n 2) part is confusing me, ive always been quite confused on questions like these =/
Reply 8
Original post by TimetoSucceed
If you expand that wouldn't you just get 1+e^jntheta, im finding this very difficult


No. You should review the binomial expansion.
Reply 9
Original post by TimetoSucceed
If you expand that wouldn't you just get 1+e^jntheta, im finding this very difficult


No - what does the binomial theorem tell you about the expansion of (1+z)n(1+z)^n?

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