How Do You Prove the Sum of a Geometric Series in Complex Numbers?

AI Thread Summary
The discussion focuses on proving the sum of a geometric series in complex numbers, specifically the formula Σ(z^k) from k=0 to n equals (z^(n+1) - 1)/(z - 1) for z in the complex numbers. Participants suggest starting with the polynomials P1(z) = z^n - 1 and P2(z) = z - 1 to demonstrate the division of these polynomials. One user highlights the importance of rearranging the terms in the proof, noting that subtracting Sn from zSn yields the same result. The conversation also touches on the uniqueness of the solution for P2 = 0, reinforcing the division approach. Overall, the thread emphasizes a structured method to derive the geometric series sum using polynomial properties.
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Prove:

\sum^{n}_{k=0} {z^{k}} = \frac{z^{n+1} -1}{z-1}

z \in \mathbb{C}

I have no idea where to even start. Just some hints on the strategy/pattern we are supposed to see would be great.
 
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Oh yeah, I forgot:

k \in \mathbb{N}
 
HINT:Consider the polynomials
P_{1}(z)=z^{n}-1
P_{2}(z)=z-1

Can u show that ¨P_{2}/P_{1}??.Then simply divide the first through the second...

Daniel.

EDIT:Then make n->n+1 and u'll get your result...
 
Last edited:
quasar:

Thanks for the reminder! :smile: I had forgotten that derivation. It seems that if you subtract Sn from zSn instead of the other way around like they did, then you'll arrive at the answer I'm supposed to show. Obviously it doesn't matter, since both numerator and denominator would be the negative of what they were if you did it the other way 'round. Thanks.

dexter:

"Can u show that P2/P1 ?? "...is not quite a complete thought. :smile: I'm just curious as to what about them you intended for me to show.

Thanks.
 
What is the solution of the eq.P_{2}=0.Is it unique??If so,then,if P_{1}(1) = 0[/tex],it means that the second polynomial (P_{2}) divides the first.Then u can divide the first through the second...<br /> <br /> Daniel.
 

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