Complex Numbers Help needed.

I tried the following approach.

Let z=eiθ2 z = \dfrac{ e^{ i \theta }}{2} . Then using de Moivre's theorem:

n=1Nsinnθ2n= Im (n=1Nzn) \sum _ { n = 1 } ^ { N } \frac{ \sin n \theta } { 2 ^ n } = \text { Im } \left ( \sum _ { n = 1 } ^ { N } z^ { n} \right)

This is sum of GP. Using the sum of GP:

n=1Nzn=eiθ2(1eNiθ2N)1eiθ2 \sum _{ n = 1 } ^{ N } z^ { n} = \frac{ \frac{e ^{ i \theta } } { 2 } ( 1 - \frac{e ^{ N i \theta } } {2^ { N } } ) } { 1- \frac{ e^{i \theta } } { 2 } }

This simplifies to:

n=1Nzn=eiθ(2NeNiθ)2N(2eiθ) \sum _{ n = 1 } ^{ N } z^ { n} = \frac{ e ^{ i \theta } ( 2 ^ { N } - e ^{ N i \theta } ) } { 2 ^ { N} ( 2- e^{i \theta } ) }

n=1Nzn=(2Neiθe(N+1)iθ)2N(2eiθ)     () \sum _{ n = 1 } ^{ N } z^ { n} = \frac{ ( 2 ^ { N } e ^{ i \theta } - e ^{ (N + 1 ) i \theta } ) } { 2 ^ { N} ( 2- e^{i \theta } ) } ~~~~~ (*)

Now I changed the complex numbers on the RHS from Euler's form to cis form, then made the denominator real and finally took imaginary part of the expression. Is there a simpler way to solve this problem? My method is very complicated, thus error-prone.

In fact, I have already made an error which is why I have posted this note. I am getting the RHS as

2N+1sinθ2sin(N+1)θ2N(54cosθ)\frac{2 ^ { N + 1 } \sin \theta - 2 \sin (N + 1) \theta } {2 ^ N (5 - 4 \cos \theta) }

i.e. without the sinNθ\sin N \theta term in the numerator. Where is the sinNθ \sin N \theta term coming from? There are only eiθe^{ i \theta } and e(N+1)iθ e ^{ (N + 1 ) i \theta } in the ()(*) equation. I can't seem to find out where I have gone wrong.

Please help.

Thanks.

Source: CIE A Levels Further Maths May / June 2007 Q 11 (b)

Note by Pranshu Gaba
4 years ago

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I think you are missing out any term. I solved the RHS and got the desired result. Recheck it. As far as i think this is a proofing problem and the method you followed is very simple and short i think. Just need some care while solving.

Sachin Vishwakarma - 3 years, 12 months ago

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Thanks Sachin :), I realized I was silly mistakes. I solved it again carefully and now I got the desired result.

Pranshu Gaba - 3 years, 11 months ago

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take sin(x)=i2(eixeixsin(x)=\dfrac{-i}{2}(e^{ix}-e^{-ix}. show this is true by de moivres.

Aareyan Manzoor - 3 years, 11 months ago

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Thanks for helping Aareyan, this method is nice! :)

Pranshu Gaba - 3 years, 11 months ago

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