Sum of a series using complex numbers

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golgiapparatus31
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#1
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#1
https://imgur.com/a/iM1nwTU

I first write that 1 - secθ e= -i tan θ

so the sum is a geometric series with first term and common ratio equal to -i tan θ

so I apply the GP formula and simplify:

Now I have the sum being equal to
[(-i)N tanN θ - 1][itanθ + tan2θ]
divided by
[ tan2 θ +1 ]

I'm stuck here because I don't know the real part.. .. ?
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mqb2766
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N is even so ...
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DFranklin
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Why not? Multiply out the numerator and take the real part, surely?
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RDKGames
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#4
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(Original post by golgiapparatus31)
https://imgur.com/a/iM1nwTU

I first write that 1 - secθ e= -i tan θ

so the sum is a geometric series with first term and common ratio equal to -i tan θ

so I apply the GP formula and simplify:

Now I have the sum being equal to
[(-i)N tanN θ - 1][itanθ + tan2θ]
divided by
[ tan2 θ +1 ]

I'm stuck here because I don't know the real part.. .. ?
(-i)^N = \pm 1 = k since N is even.

Also, if you multiply numerator and denominator by \cos^2 \theta you should see a straight-forward reduction to the required result if you do not see it yet.
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golgiapparatus31
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Yes, I tried it but couldn't reach given result. :/ Sorry, I will try harder for the next time but I was not sure how to proceed...

The numerator is
(-1)N iN+1tanN+1 θ - i tan θ + (-i)N tanN+2 θ - tan2 θ

so if N is even 1st and 2nd term will contain i but 3rd and 4th term won't...

so Real part =
(-i)N tanN+2 theta - tan2 theta
divided by
1+tan2 theta

I multiply top/bottom by cos^2 θ

getting cos^2 θ [(-i)N tanN+2 theta - tan2 theta]

take tan^2 theta out

so I get sin^2 theta ( k tan^N theta - 1)

(Original post by RDKGames)
(-i)^N = \pm 1 = k since N is even.

Also, if you multiply numerator and denominator by \cos^2 \theta you should see a straight-forward reduction to the required result if you do not see it yet.
Thank you
I tried what you said above but I didn't reach given result ...
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RDKGames
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(Original post by golgiapparatus31)
Yes, I tried it but couldn't reach given result. :/ Sorry, I will try harder for the next time but I was not sure how to proceed...

The numerator is
(-1)N iN+1tanN+1 θ - i tan θ + (-i)N tanN+2 θ - tan2 θ

so if N is even 1st and 2nd term will contain i but 3rd and 4th term won't...

so Real part =
(-i)N tanN+2 theta - tan2 theta
divided by
1+tan2 theta

I multiply top/bottom by cos^2 θ

getting cos^2 θ [(-i)N tanN+2 theta - tan2 theta]

take tan^2 theta out

so I get sin^2 theta ( k tan^N theta - 1)


Thank you
I tried what you said above but I didn't reach given result ...
Yeah looks like a typo on the question.
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golgiapparatus31
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#7
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(Original post by RDKGames)
Yeah looks like a typo on the question.
Thank you!!

Thanks a lot everyone for your time! I greatly appreciate all the help!!!
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