Sorry for title being not full, could not write it all:
let f be analytic in $Ball_2\left(0\right)$ and f is odd.
let $U=\left\{z\in \mathbb{C}|1<\left|z\right|<2\right\}$
Prove:
$\exists f\in Analytic\left(U\right):\:\forall z\in U,\:s.t:\:z\left[1-z^2\right]F'\left(z\right)=f\left(z\right)$
My problem in this is with the solution of the professor.
He assumed I can use the given value of the proof, what I mean is:
he said that we have the $\left[1-z^2\right]F'\left(z\right)=f\left(z\right)$ as given, switch sides of the equation, then do integral on $F'(z)$, use residue theorem and reach zero ( it is easy to prove it ).
My problem with the proof is, he is using what I have to prove as given, he does not show a way to reach the equation $\left[1-z^2\right]F'\left(z\right)=f\left(z\right)$
Is the solution of the proffesor right? is it true that I can always use the data given in what I have to proof? it just does not seems right to me.
The equation $$z(1-z^2)f'(z)=f(z)$$ has no nontrivial odd solutions holomorphic in $|z|<2.$ According to my calculations if $$f(z)=\sum a_n z^{2n+1}$$ then $$a_n=a4^{-n}{2n\choose n}$$ Hence the radius of convergence is equal $1.$
However I suspect that the original task was: for a given odd function $f,$ find $F$ satisfying the equation $$z(1-z^2)F'(z)=f(z)$$ The solution is presented in my comment.