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    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?

    Thanks in advance
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    (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?

    Thanks in advance
    I worked this through very quickly and got \frac{(1-\cos 4nx)-i\sin 4nx}{2 \sin x} but I may have screwed up.

    Also Is your quoted modulus correct? I got \frac{\sin 4nx}{\sin x}}

    To find the modulus, write the complex number in the form z=a+ib, then |z|=\sqrt{a^2+b^2} but your expression for C+iS won't give you the quoted result, so something is wrong somewhere.
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    (Original post by atsruser)
    I worked this through very quickly and got \frac{(1-\cos 4nx)-i\sin 4nx}{2 \sin x} but I may have screwed up.

    Also Is your quoted modulus correct? I got \frac{\sin 4nx}{\sin x}}

    To find the modulus, write the complex number in the form z=a+ib, then |z|=\sqrt{a^2+b^2} 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.
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    (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?

    Thanks in advance
    The method is as atsruser suggested, and does work out to \frac{\sin nx}{\sin x}

    Post some working if you'd like someone to check.
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    (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.
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    (Original post by ghostwalker)
    Agreed.



    The method is as atsruser suggested, and does work out to \frac{\sin nx}{\sin x}

    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))
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    (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 C+iS= \frac{\sin 2nx}{2\sin x}+\frac{1-\cos 2nx}{2\sin x}i

    Squaring and adding C and S, we get:

    C^2+S^2= \frac{\sin^2 2nx}{4\sin^2 x}+\frac{(1-\cos 2nx)^2}{4\sin^2 x}

    = \frac{1 +1-2\cos 2nx}{4\sin^2 x}

    = \frac{1 -\cos 2nx}{2\sin^2 x}

    Then double angle formula:

    = \frac{1 -(1-2\sin^2 nx)}{2\sin^2 x}

    and I'm sure you can finish.
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    (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 C+iS= \frac{\sin 2nx}{2\sin x}+\frac{1-\cos 2nx}{2\sin x}i

    Squaring and adding C and S, we get:

    C^2+S^2= \frac{\sin^2 2nx}{4\sin^2 x}+\frac{(1-\cos 2nx)^2}{4\sin^2 x}

    = \frac{1 +1-2\cos 2nx}{4\sin^2 x}

    = \frac{1 -\cos 2nx}{2\sin^2 x}

    Then double angle formula:

    = \frac{1 -(1-2\sin^2 nx)}{2\sin^2 x}

    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!
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    (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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