If $f$ is an odd function that is holomorphic in $\mathbb{C}- \{0\}$ such that $|f(z)| \leq \frac1{|z|}+ |z|^2$, then $f(z) = \frac{a_{-1}}{z} + a_1z$

Question

Let $f$ be an odd function that is holomorphic in $\mathbb{C}- \{0\}$ such that $|f(z)| \leq \dfrac{1}{|z|}+ |z|^2, $ where $z \neq 0.$

Could someone advise on how to show $f(z) = \dfrac{a_{-1}}{z} + a_{1}z, $ where $a_{-1}, a_{1}$ are the respective coefficients of $1/z$ and $z$ in Laurent series of $f \ ?$

Hints will suffice, thank you.

Answer 1

1. Use the hint given by Kelenner.
2. Since $f$ is odd, $\mathrm{Res}(f(z)/z,0)=0$, which implies the Laurent series of $f(z)$ without constant term.

Answer 2

Thanks for all the advice.

Given $R > 0 \ $, $|g(z)| \leq 1+R^3, \forall z \in D(0,R)-\{0\}.$ Hence $g$ has removable singularity at $z=0$ and $g$ can be extended to an entire function.

In particular, for $z \neq 0, \ g(z)=\sum^{\infty}_{n=0}a_nz^n=z\left(\dfrac{a_0}{z}+a_1+a_2z+ \ ...\right).$

So, for $z \neq 0,\ f(z)=\dfrac{c_{-1}}{z}+c_0+c_1z+ \ ...,$ where $c_j=a_{k-1}$

For $ j \geq 3, \ |c_j|= \begin{align} \left|\dfrac{1}{2\pi i}\int_{C(0,R)}\dfrac{f(s)}{s^{j+1}}ds\right| \leq \dfrac{2 \pi R}{2 \pi R^{j+1}}\left(\dfrac{1}{R}+R^2\right)\end{align} \to 0,$ as $R \to \infty.$

Since $f$ is odd, $c_0=c_2=0.$ (Qed)