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Homework Help: Simple question

  1. Dec 13, 2008 #1
    Can somebody explain why

    [tex]\frac{-1}{(2 \pi )(2+in)}[e^{- \pi (2 + in)} - e^{\pi (2 + in)}][/tex]

    can be written

    [tex]\frac{(-1)^n}{2+in} \cdot \frac{e^{2 \pi} - e^{-2 \pi}}{2 \pi}[/tex]

    where n is an integer?
     
  2. jcsd
  3. Dec 13, 2008 #2

    Dick

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    E.g. e^(pi*(2+i*n))=e^(2*pi)*e^(i*pi*n). e^(i*pi*n)=cos(pi*n)+i*sin(pi*n). sin(pi*n)=0. cos(pi*n)=(-1)^n. That sort of thing.
     
  4. Dec 13, 2008 #3
    I still don't understand why it's

    [tex]\frac{(-1)^n}{2+in} \frac{e^{2 \pi} - e^{-2 \pi}}{2 \pi}[/tex]

    and not

    [tex]\frac{-(-1)^n}{2+in} \frac{e^{2 \pi} - e^{-2 \pi}}{2 \pi}[/tex]
     
  5. Dec 13, 2008 #4

    Dick

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    Because it's not. If you do it carefully you'll see that they used the (-1) to flip the order of e^(-2pi)-e^(2pi) into e^(2pi)-e^(-2pi).
     
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