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Determination of logarithm (complex)

  1. Apr 16, 2006 #1

    quasar987

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    The question is this:

    Consider p(z) a polynomial and C a closed path containing all the zeroes of p in its interior. Compute

    [tex]\frac{1}{2\pi i}\int_C z\frac{p'(z)}{p(z)}dz[/tex]

    The solution given by the manual starts by saying that

    [tex]\frac{p'(z)}{p(z)}=(log(p(z)))'[/tex].


    But there is no determination of log(p(z)) on C. Isn't that a problem?
     
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  3. Apr 16, 2006 #2

    Hurkyl

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    I imagine it would depend on what you do next.
     
  4. Apr 16, 2006 #3

    quasar987

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    Next we differentiate log(p(z)), it gives

    [tex]z\frac{p'(z)}{p(z)} = \sum_{j=1}^k \frac{zn_j}{z-zj}[/tex]

    where zj are the zeros of p(z) and nj their relative order of multiplicity.

    Then we use the residue theorem on the integral ansd use the fact that each zj is a simple pole for [itex]\sum_{j=1}^k \frac{zn_j}{z-z_j}[/itex] such that

    [tex]\sum_{j=1}^k Res(z\frac{p'(z)}{p(z)},z_j) = \sum_{j=1}^k \lim_{z\rightarrow z_j}(z-z_j)\frac{zn_j}{z-z_j} = \sum_{j=1}^k z_jn_j[/tex]

    to get the result.
     
    Last edited: Apr 16, 2006
  5. Apr 16, 2006 #4

    Hurkyl

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    So do you really need to specify a branch of log on the entire curve? You're only using it to prove an algebraic identity.
     
  6. Apr 16, 2006 #5

    quasar987

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    I hear you!
     
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