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    Helllo, I am struggling on the last part of this question, need more of a bump in the right direction that a solution if possible. Thanks in advance.

    For part 1, I got the imaginary parts to be e^(x)isin(y)
    For part 2, I got the first four terms of the series to be 1 + x + x^(2)/2 + x^(3)/6

    I do see a link between the series for e^(z) and the infinite series for both C and S but I don't know where to go from there.

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    (Original post by Quido)
    Helllo, I am struggling on the last part of this question, need more of a bump in the right direction that a solution if possible. Thanks in advance.

    For part 1, I got the imaginary parts to be e^(x)isin(y)
    For part 2, I got the first four terms of the series to be 1 + x + x^(2)/2 + x^(3)/6

    I do see a link between the series for e^(z) and the infinite series for both C and S but I don't know where to go from there.
    If you write C+iS in terms of exponentials, you should recognise a well known expansion.

    Spoiler:
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    See part (ii)
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    (Original post by Quido)

    I do see a link between the series for e^(z) and the infinite series for both C and S but I don't know where to go from there.
    What's the link you've got then?
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    (Original post by ghostwalker)
    What's the link you've got then?
    I got C + iS = 1 + e^ix + 1/2 e^i2x + 1/6 e^i3x...
    I see this as the maclaurin expansion for e^x with each x term replaced with e^ix.
    I don't know where to go from there though
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    (Original post by Quido)
    I got C + iS = 1 + e^ix + 1/2 e^i2x + 1/6 e^i3x...
    I see this as the maclaurin expansion for e^x with each x term replaced with e^ix.
    I don't know where to go from there though
    Okay so you have \displaystyle C+iS=e^{e^{i\theta}} now you can replace e^{i\theta} with \cos(\theta)+i\sin(\theta) and proceed to find the real and imaginary part of this leftover expression in the same way as in the first part of the question as they will be equal to C and S respectively.
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    (Original post by Quido)
    I got C + iS = 1 + e^ix + 1/2 e^i2x + 1/6 e^i3x...
    I see this as the maclaurin expansion for e^x with each x term replaced with e^ix.
    I don't know where to go from there though
    OK,

    So, we can say \displaystyle C+jS = e^{(e^{jx})}

    You know how to deal with an exponent of the form x+yj, so what's e^{jx} in that form?
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    (Original post by ghostwalker)
    OK,

    So, we can say \displaystyle C+jS = e^{(e^{jx})}

    You know how to deal with an exponent of the form x+yj, so what's e^{jx} in that form?
    So it would be e^{cos(\theta)+isin(\theta)}

    Would the Re just be e^{cos(\theta)} or e^{cos(\theta)}cos(sin(\theta))
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    (Original post by Quido)
    So it would be e^{cos(\theta)+isin(\theta)}

    Would the Re just be e^{cos(\theta)} or e^{cos(\theta)}cos(sin(\theta))
    The latter, e^{cos(\theta)}cos(sin(\theta)), since it's just a case of substituting for your x and y into the formula given in the first part.
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    (Original post by ghostwalker)
    The latter, e^{cos(\theta)}cos(sin(\theta)), since it's just a case of substituting for your x and y into the formula given in the first part.
    Alright, thank you very much for the help!
 
 
 
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