What's with the lower bound of Laurent series?

Question

I have the simple example of finding the Laurent series of:

$f(z)=\frac{1}{z-2}$

on $|z|\leq2$ and $|z|\geq2$

Procedure to find it is rather straight forward, develop it as a Power series (in the form $\sum_0^\infty a_nz^n$ when $|z|\leq 2$:

\begin{equation} \frac{1}{z-2}=-\frac{1}{2-z}=-\frac{1}{2}\frac{1}{1-\frac{z}{2}}=\sum_{n=0}^\infty(-1)^n\frac{z^n}{2^{n+1}} \end{equation}

Then for the case of $|z|\geq2$, we develop it as a Laurent series, and we seek the form $\sum_{n=-\infty}^\infty\frac{b_n}{z^n}$:

\begin{equation} \frac{1}{z-2}=\frac{1}{z}\frac{1}{1-\frac{2}{z}}=\sum_{n=-\infty}^\infty\frac{2^{n-1}}{z^n} \end{equation}

But then it turns out the lower bounds is not supposed to be $n=-\infty$, but $n=1$. I can understand that the Power series has a lower bound of 0, but the Laurent series has a lower bound of $-\infty$ in the formula, so why is it not this, but $n=1$?

Thanks

Answer 1

The general form of a Laurent series is $\sum_{n=-\infty}^\infty b_n z^n$, but this doesn't mean that each coefficient $b_n\neq0$. Using your example, for $|z|>2$, we have $$\frac{1}{z-2}=\frac{1}{z}\cdot\frac{1}{1-(2/z)}=\frac{1}{z}\sum_{n=0}^\infty \left(\frac{2}{z}\right)^n$$

Rewriting the series so that $b_n$ is more easily identified, we have $$\sum_{n=0}^\infty 2^n\left(\frac{1}{z}\right)^{n+1}=\sum_{n=1}^\infty 2^{n-1} z^{-n}$$

Observe that this series accounts for the negative integers, so another way to write this sum (though seemingly less aesthetically pleasing) is $$\sum_{-\infty}^{-1} 2^{-n-1}z^n$$

At this point, we can define $b_n$ in the following way: $$b_n=\begin{cases} 2^{-n-1}&n\leq-1\\ 0& n>-1\end{cases}$$ It follows from this definition for $b_n$ that $$\frac{1}{z-2}=\sum_{-\infty}^\infty b_nz^n,\quad|z|>2$$