- #1
xman
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I am trying to figure out this old homework problem I haven't been able to solve. The problem goes like this:
Let f(z)=c_{00}+c_{10}x+c_{01}y+\cdots + c_{nm}x^{n}y^{m} 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 f(z)=a_0+a_1 z+\cdots +a_n z^n where a_k=\frac{1}{k!} \frac{\partial^k f}{\partial x^k} (0)
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 f^{(n)}(z)=0 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 f^{\prime}(z) =f_{x}=-i f_y where the subscripts on f represent derivatives with respect to that variable and f is assumed of the form f=u+i\,v
My thoughts were, if I take say m+1 derivatives of f, then we satisfy f^{(m+1)}(z)=0 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 z, \overline{z} but the cancellation of the \overline{z} 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.
Let f(z)=c_{00}+c_{10}x+c_{01}y+\cdots + c_{nm}x^{n}y^{m} 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 f(z)=a_0+a_1 z+\cdots +a_n z^n where a_k=\frac{1}{k!} \frac{\partial^k f}{\partial x^k} (0)
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 f^{(n)}(z)=0 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 f^{\prime}(z) =f_{x}=-i f_y where the subscripts on f represent derivatives with respect to that variable and f is assumed of the form f=u+i\,v
My thoughts were, if I take say m+1 derivatives of f, then we satisfy f^{(m+1)}(z)=0 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 z, \overline{z} but the cancellation of the \overline{z} 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.
