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Complex Analysis Proof showing that a Polynomial is linear

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose P is a polynomial such that P(z) is real iff. z is real. Prove that P is linear.

    The hint given in the text is to set P = u + iv, z = x+iy and note that v = 0 iff y = 0.

    We are then told to conclude that
    a. either v-sub y(partial of v with respect to y) is greater than or equal to 0 throughout the real axis or is less than or equal to 0 throughout the real axis;

    b. either u-sub x(partial of u with respect to x) is greater than or equal to zero or is less than or equal to zero for all real values and hence u is monotonic along the real-axis;

    c. P(z) = alpha has only one solution for real-valued alpha.


    2. Relevant equations
    Cauchy-Reimann equations.
    Fundamental Theorem of Algebra (FTA) (i think)


    3. The attempt at a solution

    My friend was of the opinion that we can say v > 0 for y > 0, then we must have v < 0 for y < 0 because the FTA says v greater than or equal to 0 for all z is impossible. (yet i do not see how this follows at all and i am not sure the path to a solution that this will bring.

    I am not quite sure how the proof is supposed to follow from the given guidelines and any help would be much appreciated (even a hint).
     
  2. jcsd
  3. Feb 10, 2009 #2
    anyone have any ideas? after thinking some more i realize that b will follow from cauchy reimann equations so long as i have a. but, i fail to see how a, b, and c together define the polynomial P(z) to be linear or how part a is true.
     
  4. Feb 10, 2009 #3

    Office_Shredder

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    Hmmm... I'm not sure off the top of my head, but notice P(z) being real when z is real isn't a very enlightening statement, as any real polynomial satisfies this. So the meat of the proof must lie in P(z) being real only when z is real
     
  5. Feb 10, 2009 #4

    Dick

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    It's really not clear to me how to prove your hint a). But here's another approach. First show P(z) is a polynomial in z with real coefficients. Now from your knowledge of real polynomials, there is a real number A such that P(x)=A has either one or zero REAL roots (depending on whether the highest power of x is odd or even. That means if it's degree is greater than one, it must have a complex root r. Hence?
     
  6. Feb 10, 2009 #5
    Thanks a lot Dick I was able to put together what you said with some of my professor's help to solve the problem. he didn't like the books method either. so i had to show that the power of the polynomial couldn't be even or an odd number that is three or greater. hence, it must be one.
     
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