# Finding a Laurent series for a complex function with two poles where one is outside of the region

when finding the Laurent series for $$\frac{1}{(z-1)(z-2)}$$ in the region $$|z| <1$$, I thought that one should first evaluate where the poles are, inside or outside the given region. z=2 is outside the region, but still, Saff and Snyder include that term in the Laurent series obtaining:

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

Such that they get for $$|z|<1$$ the following series for the first term:

$$$$\frac{1}{z-2}=-\frac{1}{2}\frac{1}{1-\frac{z}{2}}=\frac{1}{2}\sum_{n=0}^{\infty}\big(\frac{z}{2}\big)^n=-\sum_{n=0}^{\infty}\big(\frac{z^n}{2^{n+1}}\big)$$$$

for the second term they get:

$$\frac{1}{z-1}= -\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$

which they combine by subtraction to get :

$$$$\sum_{n=0}^{\infty}\bigg(-\frac{1}{2^{n+1}}+1\bigg)z^n$$$$

But for another case $$\frac{z^2-2z+3}{z-2}$$, Saff and Snyder write "Notice that the region $$|z-1| <1$$ excludes $$z=2$$."

So isn't this as the residue theorem, where poles outside of the region are simply not included in the calculation - hence, one should not include them in the Laurent series?

This was confusing indeed.

• $z=1$ is also outside the region. There are no poles. Jan 3, 2022 at 11:31
The function is analytic in $$|z| <1$$ and its power series expansion is its Laurent series also. So the correct answer is $$\sum (1-2^{-n-1})z^{n}$$ [You missed $$z^{n}$$ in the series]. Note that poles and risidues play no role in this.