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Proof of complex sum

  1. Jan 15, 2005 #1

    cepheid

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    Prove:

    [tex] \sum^{n}_{k=0} {z^{k}} = \frac{z^{n+1} -1}{z-1} [/tex]

    [tex] z \in \mathbb{C} [/tex]

    I have no idea where to even start. Just some hints on the strategy/pattern we are supposed to see would be great.
     
  2. jcsd
  3. Jan 15, 2005 #2

    cepheid

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    Oh yeah, I forgot:

    [tex] k \in \mathbb{N} [/tex]
     
  4. Jan 15, 2005 #3

    dextercioby

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    HINT:Consider the polynomials
    [tex] P_{1}(z)=z^{n}-1 [/tex]
    [tex] P_{2}(z)=z-1 [/tex]

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

    Daniel.

    EDIT:Then make n->n+1 and u'll get your result...
     
    Last edited: Jan 15, 2005
  5. Jan 15, 2005 #4

    quasar987

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  6. Jan 15, 2005 #5

    cepheid

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    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.
     
  7. Jan 15, 2005 #6

    dextercioby

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    What is the solution of the eq.[itex] P_{2}=0 [/itex].Is it unique??If so,then,if [itex] 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...

    Daniel.
     
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