Consider a Laurent series $\sum_{n=-\infty}^{+\infty} a_n(z-z_0)^n$, where $a_n,z,z_0 \in \mathbb{C}$.
Is it true that $\sum_{n=-\infty}^{+\infty} a_n(z-z_0)^n=\sum_{n=0}^{+\infty} \left(a_n(z-z_0)^n+\frac{b_n}{(z-z_0)^n}\right)$ where $b_n=a_{-n}$?
I know that by definition $$\sum_{n=-\infty}^{+\infty} a_n(z-z_0)^n=\sum_{n=0}^{+\infty} a_n(z-z_0)^n+\sum_{n=1}^{+\infty} \frac{b_n}{(z-z_0)^n}.$$
However, let's consider the following example:
$\sum_{n=0}^{+\infty} (n+0i)$ diverges, $\sum_{n=0}^{+\infty} (-n+0i)$ diverges,
but $\sum_{n=0}^{+\infty} [(n+0i)+(-n+0i)]=\sum_{n=0}^{+\infty} 0$ converges.
So, based on the example, I think that in general:
$$\sum_{n=-\infty}^{+\infty} a_n(z-z_0)^n \neq \sum_{n=0}^{+\infty} \left(a_n(z-z_0)^n+\frac{b_n}{(z-z_0)^n}\right).$$
But this implies even that $$\sum_{n=0}^{+\infty} a_n(z-z_0)^n + \sum_{n=1}^{+\infty} \frac{b_n}{(z-z_0)^n} \neq \sum_{n=0}^{+\infty} \left(a_n(z-z_0)^n+\frac{b_n}{(z-z_0)^n}\right), $$
and this is not true, I guess.
Thank you!
Yes, in general they are different, but that's just because the term $a_0$ appears twice in $\displaystyle\sum_{n=0}^\infty\left(a_n(z-z_0)^n+\frac{b_n}{(z-z_0)^n}\right)$ and only once in $\displaystyle\sum_{n=-\infty}^\infty a_n(z-z_0)^n$; if $a_0=0$, they are actually equal.
On the other hand, we have $\displaystyle\sum_{n=-\infty}^\infty a_n(z-z_0)^n=a_0+\sum_{n=1}^\infty\left(a_n(z-z_0)^n+\frac{b_n}{(z-z_0)^n}\right)$. This follows from the fact that the convergence is absolute in the annulus of convergence.