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Complex analysis - the logarithmic function

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the function Log(-z) + i(pi) is a branch of logz analytic in the domain D* consisting of all points in the plane except those on the nonnegative real axis.

    2. Relevant equations



    3. The attempt at a solution

    I know that log z: = Log |z| + iArgz + i2k(pi)
    I'm not sure where to start with this question, any help would be greatly appreciated!
    thanks :)
     
  2. jcsd
  3. Oct 25, 2009 #2

    Dick

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    You know Log(z) is analytic except for a branch cut on the negative real axis. That means Log(-z) is analytic except for a branch cut on the positive real axis. So is Log(-z)+i*pi. Now you just have to show that it is a branch of log(z) by showing exp(Log(-z)+i*pi)=z.
     
  4. Oct 25, 2009 #3
    so to show that what i did was...

    exp(Log(-z) + i*pi) = exp(Log(-z))exp(i*pi) = (-z) (-1) = z

    but.. im still unclear how this shows that Log(-z) + i*pi is a branch of log z.
     
  5. Oct 25, 2009 #4

    Dick

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    If g(z) satisfies exp(g(z))=z then it's a branch of log(z). That's what defines log(z). It's an inverse function of exp(z).
     
  6. Oct 25, 2009 #5
    oh ok i see now!
    Thanks a lot!
     
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