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# FP2 - Complex Exponents watch

1. Hello,
There are two series C = cosx + cos3x + cos5x +...+cos(2n-1)x
and S = sinx + sin3x + sin5x +...+sin(2n-1)x

I have summed the series and got C + iS = (cos2nx - 1 + isin2nx)/2isinx
(Not sure if that is right but looks like it?)

How would I find the modulus of C + iS? The answer is sin(nx)/sinx but I don't see how to get there which is why I'm unsure about getting the sum of the series correct. Also how would I find the argument of C + iS?

2. (Original post by Quido)
Hello,
There are two series C = cosx + cos3x + cos5x +...+cos(2n-1)x
and S = sinx + sin3x + sin5x +...+sin(2n-1)x

I have summed the series and got C + iS = (cos2nx - 1 + isin2nx)/2isinx
(Not sure if that is right but looks like it?)

How would I find the modulus of C + iS? The answer is sin(nx)/sinx but I don't see how to get there which is why I'm unsure about getting the sum of the series correct. Also how would I find the argument of C + iS?

I worked this through very quickly and got but I may have screwed up.

Also Is your quoted modulus correct? I got

To find the modulus, write the complex number in the form , then but your expression for C+iS won't give you the quoted result, so something is wrong somewhere.
3. (Original post by atsruser)
I worked this through very quickly and got but I may have screwed up.

Also Is your quoted modulus correct? I got

To find the modulus, write the complex number in the form , then but your expression for C+iS won't give you the quoted result, so something is wrong somewhere.
Yeah, I tried finding the modulus with mine and it doesn't work. By the way, what was first term and common ratio? Mine were e^ix and e^i2x respectively.
4. (Original post by Quido)
Hello,
There are two series C = cosx + cos3x + cos5x +...+cos(2n-1)x
and S = sinx + sin3x + sin5x +...+sin(2n-1)x

I have summed the series and got C + iS = (cos2nx - 1 + isin2nx)/2isinx
(Not sure if that is right but looks like it?)
Agreed.

How would I find the modulus of C + iS? The answer is sin(nx)/sinx but I don't see how to get there which is why I'm unsure about getting the sum of the series correct. Also how would I find the argument of C + iS?

The method is as atsruser suggested, and does work out to

Post some working if you'd like someone to check.
5. (Original post by Quido)
Yeah, I tried finding the modulus with mine and it doesn't work. By the way, what was first term and common ratio? Mine were e^ix and e^i2x respectively.
Yes, that was what I worked with, but I'm short of time to check details now. I'll put up my working later/tomorrow if no one else has responded with a more informative answer.
6. (Original post by ghostwalker)
Agreed.

The method is as atsruser suggested, and does work out to

Post some working if you'd like someone to check.
Working with the numerator only, I get C^2 = cos^2(2nx) - 2cos(2nx) + 1
and S^2 = -sin^2(2nx)

Adding those gives me 1 - cos(2nx)

After square rooting the denominator I am left with Root(1-cos(2nx))/(2isin(x))
7. (Original post by Quido)
Working with the numerator only, I get C^2 = cos^2(2nx) - 2cos(2nx) + 1
and S^2 = -sin^2(2nx)

Adding those gives me 1 - cos(2nx)

After square rooting the denominator I am left with Root(1-cos(2nx))/(2isin(x))
OK,

There are a few minor errors in there with the "i"s that don't really effect the outcome, but rather than correct, I'll redo.

If we work from C + iS = (cos2nx - 1 + isin2nx)/2isinx

Then

Squaring and adding C and S, we get:

Then double angle formula:

and I'm sure you can finish.
8. (Original post by ghostwalker)
OK,

There are a few minor errors in there with the "i"s that don't really effect the outcome, but rather than correct, I'll redo.

If we work from C + iS = (cos2nx - 1 + isin2nx)/2isinx

Then

Squaring and adding C and S, we get:

Then double angle formula:

and I'm sure you can finish.
Oh right, I was taking the real parts to be the cos and imaginary to be sin. Thanks for this!
9. (Original post by Quido)
Hello,
There are two series C = cosx + cos3x + cos5x +...+cos(2n-1)x
and S = sinx + sin3x + sin5x +...+sin(2n-1)x

I have summed the series and got C + iS = (cos2nx - 1 + isin2nx)/2isinx
(Not sure if that is right but looks like it?)
I now also agree - I'd managed to add in two different factors of 2, and i somewhere else. [Hmm - are you out by a factor of i though?]

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