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Problem with Complex Polynomial!

  1. Oct 26, 2006 #1
    I am trying to figure out this old homework problem I haven't been able to solve. The problem goes like this:

    Let [tex] f(z)=c_{00}+c_{10}x+c_{01}y+\cdots + c_{nm}x^{n}y^{m} [/tex] be a polynomial function of x and y. If, in addition, f is analytic function, show that f has to be a polynomial in z. Specifically, show that [tex] f(z)=a_0+a_1 z+\cdots +a_n z^n[/tex] where [tex] a_k=\frac{1}{k!} \frac{\partial^k f}{\partial x^k} (0)[/tex]

    Solution: Now I know that if f is an analytic function in an open set, then f and its derivatives of order n are also analytic. Moreover, I know that if a function is n times differentiable and [tex] f^{(n)}(z)=0[/tex] for every z in some domain where f is analytic, then f in this domain is a polynomial of degree at most n-1.

    Moreover, we also know that [tex] f^{\prime}(z) =f_{x}=-i f_y[/tex] where the subscripts on f represent derivatives with respect to that variable and f is assumed of the form [tex] f=u+i\,v[/tex]

    My thoughts were, if I take say m+1 derivatives of f, then we satisfy [tex] f^{(m+1)}(z)=0[/tex] which guarantees we have a polynomial. Then, we we can rewrite by integrating each term pairwise, since we will have constants for each integration. But, this does not lead to an expression of the form we want to show.

    So, I then tried writing f in terms of [tex] z, \overline{z}[/tex] but the cancellation of the [tex]\overline{z}[/tex] doesn't happen as I hoped and I again do not get the desired form.

    I have tried a couple other algebra tricks, but fail to get the desired form. I was hoping someone could point me in the right direction to finally solve this problem.

    Thanks in advance for any available help.
     
  2. jcsd
  3. Oct 26, 2006 #2

    Office_Shredder

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    I can't say right off the top of my head, but I definitely noticed that [tex] f^{(m+1)}(z)=0[/tex] should be [tex] f^{(n+1)}(z)=0[/tex]

    I'll think about it if nobody else comes up with an answer
     
  4. Oct 26, 2006 #3
    Thanks for taking a look!

    I don't believe it should matter, since m and n are just the order of the derivatives. So, I believe that in any three cases, i.e. m=n, m<n, and m>n, if f is analytic, and if the expressions of the higher order derivatives are of the form [tex] f^{(j)}(z)= f_{x^{(j)}} = (-1)^{j} f_{y^{(j)}}[/tex] then I believe if you can obtain derivatives with respect to either variable will eventually be zero, since derivatives of analytic functions are again analytic, then you can obtain the case where [tex] f^{(whatever)}(z)=0[/tex]
    At least this is what I am assuming.
     
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