Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook