Trig series - Complex number

**AshleyT** · March 9th 2009, 11:49 AM

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!!

**Moo** · March 9th 2009, 12:09 PM

Hello,

Originally Posted by AshleyT

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

**AshleyT** · March 9th 2009, 12:14 PM

I think i would just like to say...you are amazing! Thank-you so much!

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