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# Complex Numbers: Geometric Series Watch

1. 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 this series being equal to
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)).

If this is correct, then how do I then write this in its simplest form? (in terms of theta)

2. (Original post by Magu1re)
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 this series being equal to
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)).

If this is correct, then how do I then write this in its simplest form? (in terms of theta)

Not sure why you got negged for asking a question so I compensated for it. Anyway to the question, Your first line is fine, it's just when you sum it, you simplify it so much that it gets a bit harder to see how to convert it. Go back down your working and see if you can find somewhere where you can convert . Also check you've summed both series z + z^3 + z^5 +...+z^(2n-1) and 1/z + 1/z^3 + 1/z^5 +...+z^(2n-1) to the correct term. To convert z^2-1 in terms of theta, it should be quite clear as z = .
3. Totally did not see that this was a double posted problem!

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