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Help with complex summation further maths a levels

Please help with 9231 MJ 14 13 Q5
Reply 1
Original post by fatima_rida
Please help with 9231 MJ 14 13 Q5

It would help to post a picture of the question as well as what youve tried so far / what youre stuck with.
Reply 2
Original post by mqb2766
It would help to post a picture of the question as well as what youve tried so far / what youre stuck with.

For some reason it won't allow me to post a picture
Reply 3
Original post by fatima_rida
For some reason it won't allow me to post a picture

click on the camera icon or load it to somewhere like imgur and link the picture?
If all else fails, just type it in.
Original post by fatima_rida
Please help with 9231 MJ 14 13 Q5

9231_s14_qp_13+Q5.JPG

I think you are referring to this question.
If yes, the following should help.
z + z2 + z3 + zn is a geometric series.

To prove the sum of the series of the cosine terms, we can use de Moivre’s theorem and then consider the result of z + z2 + z3 + zn and use the real part.

If the question is incorrect, please state the following
Year:
Jun or Nov:
Paper version:
Question:
Original post by fatima_rida
For some reason it won't allow me to post a picture


You can read the following thread just in case you want to upload a picture in future to show the question or your work.
https://www.thestudentroom.co.uk/showthread.php?t=7434109
Hint z^n +1/z^n = 2cosnθ , z^n - 1/z^n = 2isin(nθ)
Original post by Eimmanuel
9231_s14_qp_13+Q5.JPG

I think you are referring to this question.
If yes, the following should help.
z + z2 + z3 + zn is a geometric series.

To prove the sum of the series of the cosine terms, we can use de Moivre’s theorem and then consider the result of z + z2 + z3 + zn and use the real part.

If the question is incorrect, please state the following
Year:
Jun or Nov:
Paper version:
Question:

Hint: z^n + 1/z^n = 2cos(nθ)
Reply 8
Original post by π/2=Σk!/(2k+1)!!
Hint: z^n + 1/z^n = 2cos(nθ)

Just for info/fun, you could try and do the question using either geometry/trig or using trig identity/telescoping(differences). Both are a bit less algebra than using a complex geometric sequence which is the way the question asks for.
Original post by mqb2766
Just for info/fun, you could try and do the question using either geometry/trig or using trig identity/telescoping(differences). Both are a bit less algebra than using a complex geometric sequence which is the way the question asks for.

If you aggressively look for conjugates the GM method comes out pretty quickly as well (obviously all the methods are "pretty much the same" underneath). I somewhat dislike the fact they've gone for a sin(A)cos(B)/sin(C) form of final answer as opposed to (cos(X)-cos(Y))/sin(C) - it doesn't feel to me like an obviously simpler answer and it obfuscates what's going on somewhat.

I'm not sure doing z^n+1/z^n helps much here, particularly given you've already done z+z^2+...+z^n for the 1st part of the question. Just take the real part of that sum (conjugates are your friend).
Original post by DFranklin
If you aggressively look for conjugates the GM method comes out pretty quickly as well (obviously all the methods are "pretty much the same" underneath). I somewhat dislike the fact they've gone for a sin(A)cos(B)/sin(C) form of final answer as opposed to (cos(X)-cos(Y))/sin(C) - it doesn't feel to me like an obviously simpler answer and it obfuscates what's going on somewhat.

I'm not sure doing z^n+1/z^n helps much here, particularly given you've already done z+z^2+...+z^n for the 1st part of the question. Just take the real part of that sum (conjugates are your friend).

Agreed about the second identity obfuscating the answer for the trig identity method, though its the natural way to write it down if you think if it as unit length sides of a (sort of) polygon and then its related to the chords/triangle sides lengths.
(edited 1 month ago)

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