Complex proof (1 Viewer)

[khorne]

z = cis(pi/n), prove 1+z+z^2 +...+z^(2n-1) = 0 and 1+z+z^2 +...+z^(n-1) = 1+icot(pi/2n)

For the first part, I said z = (-1)^(1/n) so you end up alternating between (-1)^(1/n) and (1)^(1/n) for pre n values, with n being -1 and then n+1 being -1(-1)^(1/n), so undoing.

Is it enough for me to actually just say, the terms to n-1 subtract with the terms from n+1 to 2n-1, or do I need to go through it all?

Additionally, what about the second one? Idk how to set it out really...

 

[Trebla]

I have no idea what you did but here's my solution using simple geometric series

 

[khorne]

lol I just got the geometric series as I was opening this to check for replies.

For the first one, my alternative method was:
z = (cis(pi))^1/n i.e z^n = -1

so z^1 = -1
z^2 = i
z^3 = z^2 * z = - i
z^4 = z^2 * z^2 = 1

therefore every 4 terms = 0, i.e 2n-2 would be a divisor of 4, and such z+...+z^(2n-2) = 0, z^(2n-1) = -1, thus 1 + 0 -1 = 0

 

[Trebla]

lol I just got the geometric series as I was opening this to check for replies.

For the first one, my alternative method was:
z = (cis(pi))^1/n i.e z^n = -1

so z^1 = -1
z^2 = i
z^3 = z^2 * z = - i
z^4 = z^2 * z^2 = 1

therefore every 4 terms = 0, i.e 2n-2 would be a divisor of 4, and such z+...+z^(2n-2) = 0, z^(2n-1) = -1, thus 1 + 0 -1 = 0

That doesn't make sense.
For example, how does z = -1 and then z² = i? It's already a contradiction.
z = cis(pi/n), not - 1
Keep in mind that n is a fixed number (which is currently unknown).

 

[khorne]

That doesn't make sense.
For example, how does z = -1 and then z² = i? It's already a contradiction.
z = cis(pi/n), not - 1
Keep in mind that n is a fixed number (which is currently unknown).

That makes sense. I see it. Alright, thanks. I've been all over the place today.