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How many zeros inside the disk?

  1. Feb 3, 2009 #1
    How many zeros inside the unit disk does the following function have?

    f(z) = 3 z^621 - e^z. (in words, three times z to the power 621 minus e^z :)

    ...argument principle answers this question, but I have a problem evaluating the integral of
  2. jcsd
  3. Feb 3, 2009 #2
  4. Feb 3, 2009 #3
    In general,
    [tex]\frac{d}{dz} \ln f(z) = \frac{f'(z)}{f(z)}.[/tex]
  5. Feb 4, 2009 #4


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    Where, exactly, are you having trouble?
  6. Feb 4, 2009 #5
    sure. and how does it help?
  7. Feb 4, 2009 #6
    well, f'(z)/f(z) = (1863*z^620 - e^z}) / (3*z^621 - e^z).

    Now, we need to integrate that along the contour C. I've tried to integrate it directly by setting z = e^{theta*i) where 0 < theta <= 2pi , but got stuck.
  8. Feb 4, 2009 #7
  9. Feb 4, 2009 #8
    I thought this was hint enough. An antiderivative of [tex]f'(z)/f(z)[/tex] is [tex]\ln f(z) = \ln(3z^{621} - e^z)[/tex].

    Am I missing something?
  10. Feb 4, 2009 #9
  11. Feb 4, 2009 #10
    Maybe the argument principle isn't what you should be using?
  12. Feb 4, 2009 #11
    do you know how else I can solve this problem?
  13. Feb 5, 2009 #12
    I cheated a bit and made some plots, but it looks like 3zn - ez has n zeros in the unit disk, and they look like they're nearish to the zeros of zn - 1. It looks like you have to use the fact that e < 3, since replacing the constant 3 with any number greater than e gives the same number of zeros. If you replace it with e, then you get a zero at z = 1.
  14. Feb 5, 2009 #13
  15. Feb 5, 2009 #14
    Looks like that does it. :)

    I just started taking a complex analysis course, so unfortunately I'm not very well-informed with regards to that stuff. Glad you solved it, though.
  16. Feb 5, 2009 #15
    cheers. now I can move on relieved :)
  17. Feb 5, 2009 #16


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    For the record, adriank's earlier hint would have let you directly compute the contour integral... except for the sticky problem that the value of the integrand wraps around the origin (621 times!), and thus must cross a branch cut (621 times), so the straightforward thing wouldn't work.

    Of course, if you could prove that it wraps around the origin exactly 621 times, and thus crosses the branch cut 621 times, you could take advantage of the fact you know exactly how big the discontinuity is, and so you still could make a direct calculation of the contour integral.
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