Complex analysis help! Watch

Idg a damn
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Please anyone help me!
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DFranklin
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What have you done so far?
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Idg a damn
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bump
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Idg a damn
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anyone?????
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Idg a damn
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at least gimme a hint
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RDKGames
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(Original post by Idg a damn)
at least gimme a hint
Note that z = e^{i \theta} = \cos \theta + i \sin \theta. So z^{2n-1} = \cos[(2n-1)\theta] + i \sin[(2n-1)\theta] by De Moivre's Theorem.

Therefore \displaystyle \sum_{n=1}^N z^{2n-1} = \sum_{n=1}^N \cos[(2n-1)\theta] + i \sum_{n=1}^N \sin[(2n-1)\theta].

So if you can work out what  \displaystyle \sum_{n=1}^N z^{2n-1} is, then just take the real part of the answer to obtain \displaystyle \sum_{n=1}^N \cos[(2n-1)\theta].

To work out that sum, you can try and think about how  \displaystyle \sum_{n=1}^N (e^{i\theta})^{2n-1} is actually just a geometric series.
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Idg a damn
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(Original post by Idg a damn)
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Is here anything I can do to reduce the numerator into two different complex numbers?
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Idg a damn
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(Original post by RDKGames)
Note that z = e^{i \theta} = \cos \theta + i \sin \theta. So z^{2n-1} = \cos[(2n-1)\theta] + i \sin[(2n-1)\theta] by De Moivre's Theorem.

Therefore \displaystyle \sum_{n=1}^N z^{2n-1} = \sum_{n=1}^N \cos[(2n-1)\theta] + i \sum_{n=1}^N \sin[(2n-1)\theta].

So if you can work out what  \displaystyle \sum_{n=1}^N z^{2n-1} is, then just take the real part of the answer to obtain \displaystyle \sum_{n=1}^N \cos[(2n-1)\theta].

To work out that sum, you can try and think about how  \displaystyle \sum_{n=1}^N (e^{i\theta})^{2n-1} is actually just a geometric series.
I have already considered the sum to be a geometric series, I just need help in reducing the expression into what they want.
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Idg a damn
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Is there anything I could use? A trig identity? Or something?
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RDKGames
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(Original post by Idg a damn)
Is there anything I could use? A trig identity? Or something?
I would really rather stick to summing up in terms of e^{i\theta}, and then at the end use the fact that 2i\sin \theta = e^{i\theta} - e^{-i\theta}
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Idg a damn
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Managed to solve it just now, no thanks to you ( I wasn't being sarcastic nor was I trying to be offensive )
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