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Branch points

  1. Aug 25, 2011 #1
    Hi everyone,

    Could someone please help me calculate the branch points?

     
  2. jcsd
  3. Aug 25, 2011 #2

    HallsofIvy

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    The "branch point" of ln(z) itself is z= 0. So you need to solve [itex]z^2+ 2z+ 3= 0[/itex].
     
  4. Aug 29, 2011 #3
    I did. My answer is [tex]-1 \pm i[/tex] but Wolfram Alpha gives it as [tex]-1 \pm \sqrt{2} i[/tex]
     
  5. Aug 29, 2011 #4

    HallsofIvy

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    Well, since you don't say how you got [itex]-1\pm i[/itex], I can only say that Wolfram Alpha is correct.
     
  6. Aug 29, 2011 #5
    I'm sorry, I made a stupid mistake with my quadratic formula. I now have the same answer as Wolfram.

    So do I basically substitute the values back into the expression within the logarithm? In that case, I get [tex]\log(2-\sqrt{2})[/tex] and [tex]\log(2+\sqrt{2})[/tex]
     
  7. Aug 29, 2011 #6
    No. You need do nothing more to identify the branch points and sides, if you back-substituted the zeros of that quadratic back into the quad, you should get zero.
     
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