How to find the Laurent series of $\frac{1}{(z - 1) (z - 2)}$ at $z=1$?

Question

This is what I did $$ \begin{align} \frac{1}{(z - 1) (z - 2)} &= -\frac{1}{ (z - 1) (1 - (z - 1))}\\ & = \frac{1}{ (z - 1)^2} \left(\frac{1}{1-\frac{1}{(z - 1)} }\right) \\ &= \frac{1}{ (z - 1)^2} \left(1 + \frac{1}{(z - 1)} + \frac{1}{(z - 1)^2} + \frac{1}{(z - 1)^3}\,+\,...\right) \\ &= \frac{1}{(z - 1)^2}+\frac{1}{(z - 1)^3}+\frac{1}{(z - 1)^4}+\frac{1}{(z - 1)^5}+\,... \end{align} $$

but my teacher said it is wrong and said that this is the right expansion

$$ \begin{align} \frac{1}{(z - 1) (z - 2)} &= -\frac{1}{ (z - 1) (1 - (z - 1))}\\ & = -\frac{1}{ (z - 1)} \left(\frac{1}{1 - (z - 1)}\right) \\ &= -\frac{1}{ (z - 1)} (1 + (z - 1) + (z - 1)^2 + (z - 1)^3\,+\,...) \\ &=- \frac{1}{z-1}-1-(z-1)-(z-1)^2\,-\,... \end{align} $$

My question is why is my expansion wrong?

Thank you.

Answer 1

Your attempted expansion of $\frac{1}{1 - r}$ is valid for $|r| < 1$. Now, with $z$ near $1$ and $r = \frac{1}{z-1}$, I don't think you'll be in the correct range of values for $r$...

Answer 2

The ranges of $z$ are important. They can be given here as you use the geometric series which is well known. Your expansion of $\frac{1}{1-\frac{1}{(z - 1)}}$ is valid for $|z - 1| > 1$ whereas your teacher's expansion of $\frac{1}{1-{(z - 1)}}$ is valid for $|z - 1| < 1$. Since you want $z$ to be "near 1" your teacher's expansion is the right one.