How Do We Know How To Expand Function As Laurent Series

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I'm a little confused about a one part of a specific example of a Laurent series that was given by a prof. It seems like it should be pretty straightforward. I want to expand $f(z)=\frac{1}{z^2-z}$ as a Laurent series about $z=i$. I did it one way, and he did it another.

It's analytic at this point, so I think the Laurent series should just be in the form of a regular power series expansion. We know that $f(z)=\frac{1}{z-1}+\frac{1}{z}$. I agree with the example that $\frac{1}{z-1}=\frac{1}{i-1}\cdot\frac{1}{1-(-\frac{z-i}{i-1})}=\sum_{n\geq 0} \frac{(-1)^n}{(i-1)^n}(z-i)^n$. Here is where I'm confused. I would have written $\frac{1}{z}$ as: $$ \frac{1}{z}=\frac{1}{i+(z-i)}=\frac{1}{i}\cdot\frac{1}{1-(\frac{z-i}{-i})}=\sum_{n\geq0}\frac{(-1)^n}{i^{n+1}}(z-i)^n $$

No principal part, which makes sense to me.

The prof instead did: $$ \frac{1}{(z-i)+i}=\frac{1}{z-i}\cdot\frac{1}{1-\frac{-i}{z-i}}=\sum_{n\geq 0}(-i)^n(z-i)^{-(n+1)} $$

This seems to be giving us a principal part for the Laurent expansion and a non-zero residue. Can someone explain why he did this?

Thanks!

EDIT: There was a nice comment pointing out that the series I wrote and the series the prof wrote have two different domains of convergence. I'm a little confused in general then what the prof's expansion is saying about our function. Does it matter that it has an infinite principal part? What does that mean?

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To expand on my original comment, let's investigate the domain of convergence of each series. First, we have

$$\frac{1}{z-1} = \sum_{n=0}^\infty (-1)^n \left(\frac{z-i}{i-1}\right)^n$$

which converges for $|z-i|<\sqrt{2}$. For $\frac{1}{z}$, we have two different candidates for our series. You have

$$\frac{1}{z} = \frac{1}{i} \sum_{n=0}^\infty (-1)^n \left(\frac{z-i}{i}\right)^n$$

which converges for $|z-i|<1$. Your professor has

$$\frac{1}{z} = \frac{1}{z-i}\sum_{n=0}^\infty (-1)^n \left(\frac{i}{z-i}\right)^n$$

which converges for $|z-i|>1$.

Putting each of these together, you have a Taylor series for $f$ which converges for $|z-i|<1$ (which can indeed, if you wish, be interpreted as a series defined on the punctured disc), whereas your professor has a Laurent series for $f$ which converges in the annulus $1<|z-i|<\sqrt{2}$. Perhaps your professor intended for you to expand your function in some annulus centered at $i$, other than the punctured disc?