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Complex analysis

  1. May 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose f is entire and there exist constants a and b such that ##|f(z)| \le a|z|+b## for all ##z \in C##. Prove that f is a polynomial of degree at most 1.

    2. Relevant equations

    3. The attempt at a solution
    We have that for any ##z \neq 0##, ##\frac{|f(z)|}{a|z|} \le b##. So if we take the limit as ##z \to \infty## it is obvious that if f is polynomial, it can't have a degree greater than 1. However I am not sure why it must be polynomial.
  2. jcsd
  3. May 24, 2017 #2


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    Staff Emeritus
    Science Advisor

    I'm not sure what you are supposed to be able to use in your proof. One fact about entire functions is the Cauchy estimate formula, which says:
    • Take a circle in the complex plane of radius [itex]R[/itex] centered on [itex]z=0[/itex].
    • Let [itex]M_R[/itex] be the largest value of [itex]|f(z)|[/itex] on the circle.
    • Then [itex]|f^{(n)}(0)| \leq \frac{n! M_R}{R^n}[/itex] (where [itex]f^{(n)}[/itex] means the [itex]n^{th}[/itex] derivative).
    See if you can use this to prove [itex]f^{(n)} = 0[/itex] for [itex]n > 1[/itex].
  4. May 24, 2017 #3


    Staff: Mentor

    This doesn't follow from ##|f(z) | \le a|z| + b##
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