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Math Help - Trig series - Complex number

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    Trig series - Complex number

    Urg, really struggling with this one! Any help i'd really appreciate!

    Show that(where Q is theta =) )
    <br />
\frac{e^{nQi} - 1}{e^{Qi} - 1} = \frac{sin(\frac{Qn}{2})}{sin(\frac{Q}{2})}e^\frac{  (n-1)Qi}{2}<br />
    provided that Q is not a multiple of 2pi...

    Thanks!!
    Last edited by mr fantastic; March 9th 2009 at 12:05 PM. Reason: Fixed some latex
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    Hello,
    Quote Originally Posted by AshleyT View Post
    Urg, really struggling with this one! Any help i'd really appreciate!

    Show that(where Q is theta =) )
    <br />
\frac{e^{nQi} - 1}{e^{Qi} - 1} = \frac{sin(\frac{Qn}{2})}{sin(\frac{Q}{2})}e^\frac{  (n-1)Qi}{2}<br />
    provided that Q is not a multiple of 2pi...

    Thanks!!
    You want to go from exponentials to sines.
    So you should remember this formula : \sin x=\frac{e^{ix}-e^{-ix}}{2i}
    but here, you have things like e^{\dots}-1

    so try to see what happens if you multiply the sine formula by e^{ix}, to get 1 on the right side :

    \sin(x) e^{ix}=\frac{e^{2ix}-1}{2i}
    and finally, e^{2ix}-1=2i \sin(x) e^{ix}
    which is equivalent to e^{it}-1=2i \sin(t/2) e^{it/2} (t=2x)


    Applying this to the formula you have :
    \frac{e^{nQi} - 1}{e^{Qi} - 1}=\frac{2i \sin \left(\tfrac{nQ}{2}\right) e^{inQ/2}}{2i \sin \left(\tfrac{Q}{2}\right) e^{iQ/2}}
    note that 2i can be simplified :
    =\frac{\sin \left(\tfrac{nQ}{2}\right) e^{inQ/2}}{\sin \left(\tfrac{Q}{2}\right) e^{iQ/2}}

    now using simple property of exponents, we have \frac{e^{inQ/2}}{e^{iQ/2}}=e^{\frac{(n-1)Qi}{2}}

    and this finishes the problem :P
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    I think i would just like to say...you are amazing! Thank-you so much!
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