# Sum of a series using complex numbers

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https://imgur.com/a/iM1nwTU

I first write that 1 - secθ e

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)

divided by

[ tan

I'm stuck here because I don't know the real part.. .. ?

I first write that 1 - secθ e

^{iθ }= -i tan θso 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}tan^{N}θ - 1][itanθ + tan^{2}θ]divided by

[ tan

^{2}θ +1 ]I'm stuck here because I don't know the real part.. .. ?

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#4

(Original post by

https://imgur.com/a/iM1nwTU

I first write that 1 - secθ e

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)

divided by

[ tan

I'm stuck here because I don't know the real part.. .. ?

**golgiapparatus31**)https://imgur.com/a/iM1nwTU

I first write that 1 - secθ e

^{iθ }= -i tan θso 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}tan^{N}θ - 1][itanθ + tan^{2}θ]divided by

[ tan

^{2}θ +1 ]I'm stuck here because I don't know the real part.. .. ?

Also, if you multiply numerator and denominator by you should see a straight-forward reduction to the required result if you do not see it yet.

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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)

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

so Real part =

(-i)

divided by

1+tan

I multiply top/bottom by cos^2 θ

getting cos^2 θ [(-i)

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 ...

The numerator is

(-1)

^{N}i^{N+1}tan^{N+1}θ - i tan θ + (-i)^{N}tan^{N+2}θ - tan^{2}θso if N is even 1st and 2nd term will contain i but 3rd and 4th term won't...

so Real part =

(-i)

^{N}tan^{N+2}theta - tan^{2}thetadivided by

1+tan

^{2}thetaI multiply top/bottom by cos^2 θ

getting cos^2 θ [(-i)

^{N}tan^{N+2}theta - tan^{2}theta]take tan^2 theta out

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

(Original post by

since is even.

Also, if you multiply numerator and denominator by you should see a straight-forward reduction to the required result if you do not see it yet.

**RDKGames**)since is even.

Also, if you multiply numerator and denominator by you should see a straight-forward reduction to the required result if you do not see it yet.

I tried what you said above but I didn't reach given result ...

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#6

(Original post by

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)

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

so Real part =

(-i)

divided by

1+tan

I multiply top/bottom by cos^2 θ

getting cos^2 θ [(-i)

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 ...

**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}i^{N+1}tan^{N+1}θ - i tan θ + (-i)^{N}tan^{N+2}θ - tan^{2}θso if N is even 1st and 2nd term will contain i but 3rd and 4th term won't...

so Real part =

(-i)

^{N}tan^{N+2}theta - tan^{2}thetadivided by

1+tan

^{2}thetaI multiply top/bottom by cos^2 θ

getting cos^2 θ [(-i)

^{N}tan^{N+2}theta - tan^{2}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 ...

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reply

(Original post by

Yeah looks like a typo on the question.

**RDKGames**)Yeah looks like a typo on the question.

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

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