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How would I count the zeros of zsin(z)-1 in a complex disc?

  1. Mar 27, 2010 #1
    The disc in question is {z: |z|<(n+1/2)pi}. I can't figure out how to apply Rouche to this. Any help would be appreciated.

    (This is in the context of showing all roots of zsin(z)=1 are real. I counted the zeros of zsin(z)-1 on the real axis and got 2n+2, and now I hope to get the same answer via the disc...)
     
  2. jcsd
  3. Mar 27, 2010 #2
    Does this work:

    f(z) = z sin(z) - 1

    g(z) = -R^2 sin(pz)/z

    where R is the radius of the disc and p some carefully chosen number very close to 1 and you try to show that:

    |f(z) - g(z)| < |f(z)| on the boundary of the disk?
     
    Last edited: Mar 27, 2010
  4. Mar 27, 2010 #3
    At first glance, your g(z) has an odd number of roots within the disc, whereas f(z) would have an even number (I'm fairly sure of this, since f(z)+1 is even and the first peak's height is greater than 1 - look at a plot of it on wolframalpha or something if you're not convinced). So, I don't think your function can work.

    What was your thinking behind selecting it? What exactly did you mean by 'carefully chosen p'?
     
  5. Mar 28, 2010 #4
    Yes, it seems that this doesn't work. I'll take a look later at this again.
     
  6. Mar 30, 2010 #5
    f(z) = z sin(z) - 1

    g(z) = z sin(z)

    should work if you take into account that g(z) has a double zero at z = 0 that f doesn't have.
     
  7. Mar 30, 2010 #6
    Ah! I forgot about that double zero. Very much appreciated
     
  8. May 18, 2012 #7
    So how can we apply Rouche to this? Which pair of functions do we choose?
     
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