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    I have been given

    1/2pi*[ (2pi*(-1)^n)/-i*n)]= (i*(-1)^n)/n

    However I just get the 2pi cancelling each other out and getting

    ((-1)^n)/(i*n)

    This is surely just an error in the sheet right or am I missing something?
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    (Original post by anonstudent1)
    I have been given

    1/2pi*[ (2pi*(-1)^n)/-i*n)]= (i*(-1)^n)/n

    However I just get the 2pi cancelling each other out and getting

    ((-1)^n)/(i*n)

    This is surely just an error in the sheet right or am I missing something?
    What're you trying to find??

    Are you trying to rearrange the first on into the second?
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    (Original post by TheIrrational)
    What're you trying to find??

    Are you trying to rearrange the first on into the second?
    Thanks for the reply

    1/2pi*[ (2pi*(-1)^n)/-i*n)] Is what I have been given. The answer on the r.h.s of the equals sign is the answer. But I don't see how its the answer.
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    Do you have

    \dfrac{1}{2 \pi}\dfrac{2 \pi (-1)^n}{(-i)^n}

    or

    \dfrac{1}{2 \pi}\dfrac{2 \pi (-1)^n}{-(i)^n}
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    (Original post by TenOfThem)
    Do you have

    \dfrac{1}{2 \pi}\dfrac{2 \pi (-1)^n}{(-i)^n}

    or

    \dfrac{1}{2 \pi}\dfrac{2 \pi (-1)^n}{-(i)^n}
    the bottom line is i*n, not to the power.
    Basically its the first question from this page
    http://math.ucsd.edu/~lni/math140/HW..._solutions.pdf

    I don't understand how they have gotten from the 4th line to the 5th
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    (Original post by anonstudent1)
    the bottom line is i*n, not to the power.
    Basically its the first question from this page
    http://math.ucsd.edu/~lni/math140/HW..._solutions.pdf

    I don't understand how they have gotten from the 4th line to the 5th
    So you have

    \dfrac{(-1)^n}{-in}

    I assume that they multiplied numerator and denominator by i

    though I have not looked at your link
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    (Original post by TenOfThem)
    So you have

    \dfrac{(-1)^n}{-in}

    I assume that they multiplied numerator and denominator by i

    though I have not looked at your link
    There answer is

    (i(-1)^n)/n

    If i multiplied the numerator and denominator by i, I get

    (i((-1)^n))/((-i^2)n)
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    (Original post by anonstudent1)
    There answer is

    (i(-1)^n)/n

    If i multiplied the numerator and denominator by i, I get

    (i((-1)^n))/((-i^2)n)
    I assume that you know what i^2 is

    So that you know what -i^2 is
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    (Original post by TenOfThem)
    I assume that you know what i^2 is

    So that you know what -i^2 is
    Well I just found out it equals -1 and it makes complete sense now. I didn't think you could just square it like another number before for some reason. Certainly makes my life easier, anyway thanks for your help!
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    (Original post by anonstudent1)
    Well I just found out it equals -1 and it makes complete sense now. I didn't think you could just square it like another number before for some reason. Certainly makes my life easier, anyway thanks for your help!
    Did you not know that i = \sqrt{-1}

    Are you self teaching?
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    (Original post by TenOfThem)
    Did you not know that i = \sqrt{-1}

    Are you self teaching?
    I do, I've known that for ages! Its just for some reason I didn't think you could just square it like another number and then it would just equal minus 1.

    And well i rarely go to lectures... so in a way lol

    I definitely should have cottoned on earlier, but when i've looked at something for so long and i'm sure they've just made a typo, i can get stuck on ridiculousy easy things.
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    (Original post by anonstudent1)
    I do, I've known that for ages! Its just for some reason I didn't think you could just square it like another number and then it would just equal minus 1.

    And well i rarely go to lectures... so in a way lol

    I definitely should have cottoned on earlier, but when i've looked at something for so long and i'm sure they've just made a typo, i can get stuck on ridiculousy easy things.
    Fair enough

    We usually rationalise the denominator so anything with i in the denominator will need to be * by something
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    (Original post by TenOfThem)
    Fair enough

    We usually rationalise the denominator so anything with i in the denominator will need to be * by something
    Thank you , ill be sure to apply that in the future!
 
 
 
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