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Homework Help: Proof that e^z is not a finite polynomial

  1. Mar 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that the analytic function e^z is not a polynomial (of finite degree) in the complex variable z.

    3. The attempt at a solution

    The gist of what I have so far is suppose it was a finite polynomial then by the fundamental theorem of algebra it must have at least one or more roots. e^z can never equal zero for hence this is a contradiction.

    Is it okay for me to apply the fundamental theorem of algebra like this or am I kind of using a bit too much machinery here?
  2. jcsd
  3. Mar 31, 2012 #2
    Your proof is wrong. Indeed: [itex]e^{2\pi i}=0[/itex], so the function DOES have a root!!
  4. Mar 31, 2012 #3
    Wait, what? [itex]e^{i2\pi} = 1[/itex]
  5. Mar 31, 2012 #4
    Woow, I'm stupid today. :cry:

    I'm sorry, your proof is alright!! I obviously need to get some sleep.
  6. Mar 31, 2012 #5
  7. Mar 31, 2012 #6

    Ray Vickson

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    Science Advisor
    Homework Helper

    That looks like too much machinery (at least for my tastes); I would rather just argue that a polynomial of degree n has (n+1)st derivative = 0 identically, and ask whether that can happen for exp(z).

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