Complex Analysis: If f is an odd function, find an expression for f. Complex Analysis
Suppose that $f$ is a holomorphic function in $\mathbb{C}\setminus\{0\}$ and satisfies $$\vert f(z) \vert \leq \vert z \vert ^{2} +\frac{1}{\vert z \vert ^{2}}$$
If f is an odd function, find an expression for f.
I thought about calculating the Laurent Serie of f, but I can't calculate Laurent's series explicitly (there is no way to calculate the coefficients), Dai thought about estimating, using the invariance of the circle to know which one cancels out. I cannot develop this question
 A: One can immediately see that for all $z\in\Bbb C^\times:=\Bbb C\setminus\{0\},$ we have $$\left|z^2f(z)\right|=|z|^2|f(z)|\le|z|^4+1.$$
As a result, the Laurent series for $g(z):=z^2f(z)$ cannot have any terms of power greater than $4,$ for if there were any, then for sufficiently large $|z|,$ the above inequality would fail to hold. Moreover, we see that $g$ is bounded in the punctured unit disk, so its Laurent series has no terms of negative power, and so is a polynomial of degree no greater than $4.$
Since $f$ is odd, then $g(-z)=(-z)^2f(-z)=z^2f(-z)=-z^2f(z)=-g(z),$ and so $g$ is odd. Hence, all terms of the Laurent series for $g$ have odd power, and so $$g(z)=az^3+bz$$ for some $a,b\in\Bbb C,$ whence $$f(z)=az+\frac{b}{z}.$$

Added: To see why all terms of the Laurent series of $g$ must have odd power, suppose $h$ is an odd function that is holomorphic in $\Bbb C^\times.$ Note that for all $z\in\Bbb C^\times,$ we have $$0=h(z)+h(-z).\tag{$\heartsuit$}$$ Now, write $$h(z)=\sum_{n\in\Bbb Z}b_nz^n,$$ and consider out the expansion for the right-hand side of $(\heartsuit).$
A: The inequality implies that for $|z|<1$
$$
|z^2f(z)|\le |z|^4+1\le 2\ .
$$
It follows from Riemann's theorem that $z=0$ is a removable singularity for $g(z):=z^2f(z)$. Thus, $g(z)=\sum_{n=0}^\infty a_nz^n$ on the punctured complex plane.
The polynomial growth
$$
|g(z)|\le |z|^4+1
$$
implies that $g$ is a polynomial (or order at most $4$) by Liouville’s theorem
So you can write
$$
z^2f(z)=a_0+a_1z+a_2z^2+a_3z^3+a_4z^4.
$$
The parity of $f$ then implies that $a_0=a_2=a_4=0$ by the fundamental theorem of algebra: $g(z)+g(-z)=0$ implies that
$$
a_0+a_2z^2+a_4z^4=0
$$
for all $z$. (But a nonzero polynomial has only finitely many roots.)
