Prove Laurent Series Expansion is Unique

Suppose that $f$ is holomorphic on $A=\{r<|z|<R\}$, where $0\le r<R\le \infty$. Suppose that there are two series of complex numbers $(a_n)_{n\in{\mathbb Z}}$ and $(b_n)_{n\in\mathbb Z}$ such that $f(z)=\sum_{n=-\infty}^\infty a_n z^n=\sum_{n=-\infty}^\infty b_n z^n$ for $z\in A$. Show that $a_n=b_n$ for all $n\in\mathbb Z$. This means that the Laurent series expansion is unique.

Hint: It suffices to show that if $f\equiv 0$, then $a_n=0$ for all $n$. Use $\sum_{n=0}^\infty a_n z^n=\sum_{n=-\infty}^{-1} -a_n z^n$ to construct a bounded entire function.

Hi everyone, I've set out to prove that the Laurent series expansion of a function is unique. I found a very short and nice proof of uniqueness here, however my problem's hint goes a different direction. Is there any reason not to favor the simple proof I linked to?

I'd like to figure out what to do with the hint given and what bounded entire function to construct. Once I find a bounded entire function, I have a feeling I will need to cite Liouville's Theorem to help me somehow, which says every bounded entire function on $\mathbb{C}$ is constant. Thanks for your help!

• It seems to me, at least, that the hint is in the same direction. This is the punchline of the other proof. – amcalde Mar 25 '15 at 3:09