# more complex variable help

• January 28th 2010, 12:52 PM
LCopper2010
more complex variable help
Show that for any z $\in$ C an Z not equal to 1 one has

1 + z +...+z^n = (1-z^(n+1))/(1-z)

and use it to show that

1 + $\sum$ from k=1 to n of cos(k $\theta$) = 1/2 + (sin((2n+1) $\theta$/2))/(2sin( $\theta$/2))
• January 28th 2010, 05:34 PM
TheEmptySet
Quote:

Originally Posted by LCopper2010
Show that for any z $\in$ C an Z not equal to 1 one has

1 + z +...+z^n = (1-z^(n+1))/(1-z)

and use it to show that

1 + $\sum$ from k=1 to n of cos(k $\theta$) = 1/2 + (sin((2n+1) $\theta$/2))/(2sin( $\theta$/2))

$\frac{1}{2}+\sum_{k=1}^{n}e^{iku}=\frac{e^{i(n+1)u }-1}{e^{iu}-1}-\frac{1}{2}=\frac{2(e^{i(n+1)u}-1)-(e^{iu}-1)}{2(e^{iu}-1)}=\frac{1}{2} \left( \frac{2e^{i(n+1)u}-e^{iu}-1}{e^{iu}-1}\right)$

Now multiply the numerator and denominaotor by $e^{-\frac{u}{2}}$ to get

$\frac{1}{2} \left( \frac{2e^{i(n+\frac{1}{2})u}-e^{i\frac{u}{2}}-e^{-i\frac{u}{2}}}{e^{i\frac{u}{2}}-e^{i\frac{u}{2}}}\right)=\frac{1}{2}\left( \frac{2(cos[(n+\frac{1}{2})u]+i\sin[(n+\frac{2}{2})u])-2\cos(\frac{u}{2})}{2i\sin(\frac{u}{2})}\right)$

Now just equate the real and immaginary parts to get your result :)