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Math Help - formula rearranged

  1. #1
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    formula rearranged

    Just wondering if it was possible for someone to show me how you get from this equation:

    ie^(it)/e^(-it)-1

    to this

    i/1-e^(-it)
    Cheers in advance! Its much appreciated.
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  2. #2
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    If you mean: is \frac{ie^{it}}{e^{-it}-1}=\frac{i}{1-e^{-it}}, the answer is no.

    Suppose \frac{ie^{it}}{e^{-it}-1}=\frac{i}{1-e^{-it}}. Then, by cancelling i we get \frac{e^{it}}{e^{-it}-1}=\frac{1}{1-e^{-it}}. By multiplying the LHS by \frac{-1}{-1}, we get \frac{-e^{it}}{1-e^{-it}}=\frac{1}{1-e^{-it}}. Finally we end up with -e^{it}=1, which is a contradiction.
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  3. #3
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    Quote Originally Posted by Karty View Post
    Just wondering if it was possible for someone to show me how you get from this equation:

    ie^(it)/e^(-it)-1

    to this

    i/1-e^(-it)
    Cheers in advance! Its much appreciated.
    hi Karty,

    you are losing a factor...

    \frac{ie^{it}}{e^{-it}-1}=\frac{-ie^{it}}{-(e^{-it}-1)}

    =\frac{-ie^{it}}{1-e^{-it}}=-e^{it}\ \frac{i}{1-e^{-it}}

    Either that, or you have a typo,

    \frac{ie^{-it}}{e^{-it}-1}=\frac{ie^{-it}e^{it}}{e^{it}\left(e^{-it}-1\right)}=\frac{i}{1-e^{it}}

    or

    \frac{ie^{it}}{e^{it}-1}=\frac{ie^{it}e^{-it}}{e^{-it}\left(e^{it}-1\right)}

    which leads to the result you've given
    Last edited by Archie Meade; April 25th 2010 at 08:31 AM.
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  4. #4
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    Quote Originally Posted by Karty View Post
    ie^(it)/e^(-it)-1
    As you show it:
    ie^(it)/e^(-it) - 1
    = ie^(it)e^(it) - 1
    = ie^(2it) - 1
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  5. #5
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    my mistake, there was an error in the formula:

    ie^(-it)/e^(-it)-1

    to this

    i/1-e^(-it)


    sorry to confuse, and thanks for taking the time to answer my qs!

    Karty
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