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!
Best Answer
Let $f(z)=\sum \limits_{n=-\infty}^{-1}{a_{n}}{z^n}+\sum \limits_{n=0}^{\infty}a_nz^n=\sum \limits_{n=1}^{\infty}\dfrac{a_{-n}}{z^n}+\sum \limits_{n=0}^{\infty}a_nz^n$
Define $g(z)$ as
\begin{align} g(z)&=\sum \limits_{n=0}^{\infty}a_nz^n, \forall z, |z|\leqslant R. \\ &=\sum \limits_{n=1}^{\infty}\dfrac{a_{-n}}{z^n}, \forall z, |z|>R \end{align}
Clearly $g(z)$ is bounded and analytic on both $|z|\leqslant R$ and $|z|>R$. So is bounded on entire domain.
Since $\sum \limits_{n=-\infty}^{-1}a_nz^n=\sum \limits_{n=1}^{\infty}\dfrac{a_{-n}}{z^n}$ for $|z|=R$, $g(z)$ is analytic on entire domain through analytic continuation. Thus $g(z)$ is bounded entire function, and $g(z)=c$, or $a_n=0$ for all $n$ except $n=0$.
Then if $f(z)\equiv 0$, $a_n=0,$ for all $n$, which means Laurent Series is unique.