Complex Numbers: Geometric Series

Hello

I am well and truly stuck on a question:

Use cos n0 = 0.5((z^n)+(z^-n)) to express cos0+cos30+cos50+...+cos(2n-1)0 (where I have used 0 to represent theta) as a geometric series in terms of z. Hence find this sum in terms of 0.

I have written the series as 0.5(((z^1)+(z^-1))+((z^3)+(z^-3)+((z^5)+(z^-5))+...+((z^(2n-1))+(z^(1-2n))). However, I cannot find a common ratio for this sequence and so do not know whether I have gone about it the wrong way.

Please could someone help me?

(P.S. Please quote me)

Write it as a series with the first term very negative:

$\frac{1}{2}(z^{1-2n}+z^{3-2n}+...+z^{-3}+z^{-1}+z^1+z^3+z^5+...+z^{2n-1})$

Write it as a series with the first term very negative:

$\frac{1}{2}(z^{1-2n}+z^{3-2n}+...+z^{-3}+z^{-1}+z^1+z^3+z^5+...+z^{2n-1})$

Sorry for bringing this up again but my last post was a bit premature. I did not follow the problem through to the end. I have
0.5((z^(1-2n))+(z^(3-2n))+...+(z^(2n-3))+(z^(2n-1))
which is equal to
0.5(((z^(1-2n))((z^(4n))-1))/((z^(2))-1))

How do I then write this in its simplest form? (in terms of theta)

So you have so far:
$\dfrac{z^{1-2n}}{2}\left(\dfrac{z^{4n}-1}{z^2-1}\right)$

Using a similar method to how you would have established $\cos n\theta = 0.5(z^n+z^{-n})$, you can work out that $i\sin n\theta = 0.5 (z^n-z^{-n})$. This is useful because $0.5 (z^n-z^{-n})=0.5(z^n-\frac{1}{z^n})=0.5\frac{z^{2n}-1}{z^n}$ so $2i\sin n\theta = \frac{z^{2n}-1}{z^n}$.

Your answer so far can be re-written as $\dfrac{1}{2}\left(\dfrac{(z^{4n}-1)/z^{2n}}{(z^2-1)/z}\right)$ and you can put this in terms of theta using the above formula.

I'll give you an idea of the thought process that I went through in answering this question. My first instinct was to try and use the formula $\cos n\theta = 0.5(z^n+z^{-n})$ and perhaps replacing n with some other number as appropriate. I noticed that $\cos n\theta = 0.5(z^n+z^{-n})=0.5(z^{2n}+1)/z^n$ which looked very similar to the terms at the top and bottom of the fraction in the sum of the g.p. but the problem here was the minus sign so I wouldn't have been able to use these workings to make a substitution. So then I thought, if the only difference here is a minus sign then perhaps I might have more luck getting a formula for the g.p. in terms of sin? So I put together a similar formula in terms of sin n theta and it turns out that I can use it to replace the terms involving z.