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Complete function

  1. Aug 28, 2011 #1
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 29, 2011 #2

    lanedance

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    can you elaborate on what the definition of a complete function is?
     
  4. Aug 29, 2011 #3
    I'm sorry , I meant entire function
     
  5. Aug 29, 2011 #4

    HallsofIvy

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    A function that is analytic on the entire complex plane.
     
  6. Aug 29, 2011 #5
    I know. So what should i do?
     
  7. Aug 29, 2011 #6
    Suppose there aren't any? Is that an option? Can that be justified?
     
    Last edited: Aug 29, 2011
  8. Aug 29, 2011 #7
    I don't know. I want to be sure before i write that there aren't...
     
  9. Aug 29, 2011 #8
    What makes you think there aren't? Not just by what I said huh? That exponent in there is a non-polynomial entire function which by Picard's first theorem, grows without bounds in the complex plane. z^2 has a pole at infinity which means it becomes unbounded too. What would f have to be to tame all that down to something that is never larger than 3?

    Also, I'm no expert and I'm not 100% sure about this.
     
  10. Aug 29, 2011 #9

    lanedance

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    what's the actual question as well? may help to write it explicitly

    my initial thoughts agree with jackmell

    f(z) must be entire, so as long as it is non-constant it must take every value in the complex plane (pretty much)

    so in really from f you need a function which:
    bounds the z^2 behaviour at large |z| by mult
    bounds the e^z behaviour at large |z| by addition
    has its behaviour at large |z| cancelled by the actions of ..*z^2 ..+e^x

    now one way to approach it may be to write
    f(z) = u(z) + iv(z)

    then maybe you can make use of the fact that if f is continuous, first derivtaives of u&v exist everywhere and satisfy Cacuhy Reimann, then f is holomorphic
     
  11. Aug 29, 2011 #10
    There's a much simpler approach. Note that if f is entire, then [itex]z^2f(z)-3+e^z[/itex] is also entire. And what do you know about bounded entire functions?
     
  12. Aug 30, 2011 #11

    lanedance

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    nice point
     
  13. Aug 30, 2011 #12
    I know it's equal to a constant (C). so f(z)=(C+3)/z^2 -(e^z)/(z^2)
    then I know that for f(z) to be entire C+3=0 and then I need to find what is the value of e^z/z^2 in 0.
    this is what i did in the first place...
     
  14. Aug 30, 2011 #13

    lanedance

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    not quite, this shows that the only possible form for f(z) is
    f(z) = (C+3)/z^2 -(e^z)/(z^2)

    now if you choose C=-3, you get
    f(z)=-(e^z)/(z^2)
    which is not analytic at z=0, so this is not a good choice

    are there any other choices for C that would remove this behaviour?
     
    Last edited: Aug 30, 2011
  15. Aug 30, 2011 #14

    lanedance

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    if so, check whether that give an entire function
    if not, you've proved there is no solution
     
  16. Aug 30, 2011 #15
    I don't see any other constant that would remove this behaviour. am i wrong?
    And as I understand from this thread e^z/z^2 isn't analytic so there are no functions..
     
  17. Aug 30, 2011 #16

    lanedance

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    i haven't work it through fully... however noting that e^{0}=1, then consider C=-2 as this gives
    [tex]f(z) = (1-e^z)\frac{1}{z^2}[/tex]

    this is the only value that gives f a chance of reasonable limiting behaviour at z=0, though I still don't think it makes it analytic there. Expanding e^z around 0

    [tex]e^z = 1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+..[/tex]
    [tex]1-e^z = -z-\frac{z^2}{2!}-\frac{z^3}{3!}+..[/tex]
    [tex](1-e^z)\frac{1}{z^2}= -\frac{1}{z}-\frac{1}{2!}-\frac{z}{3!}-...[/tex]

    so the pole order is reduced, but still there - though obvious you could do this for general C to show there is no available f
     
    Last edited: Aug 30, 2011
  18. Aug 30, 2011 #17
    Ok, thank you.
    One more thing, If I have a function and its denominator zero order and the numerator zero order are the same, then we have a removable singularity?
     
  19. Aug 30, 2011 #18

    lanedance

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    if they occur at the same z then yes, in effect it is a pole of order zero
     
  20. Aug 30, 2011 #19
    I'm talking about (1-cos(z^2))/z^4 it has 4th order zero
    and (e^z-1-z)/z^2 has 2nd order zero.
    then they both have a removable singularity, right?
     
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