laurent series of $1/(z-5)$ about $z_o=2$ I'm calculating the Laurent Series of $f=1/(z-5)$ about the point $z_{0}=2$.
The expansion I get has no principal part and is analytic within the circle $|z-2|<3$ since there are no singularities within that disk. I write $f=1/(z-5) = 1/(z-2 -3)= 1/3[(z-2)/3-1)]$, and since $(z-2)/3<1$, we can expand it into a convergent series without containing negative powers of $z-2$.
However, I saw a book say that the answer does have a principal principal part, and in their derivation they introduce a singularity at $z_{0}=2$ . I don't see why they did that. 

See if this is it.
For $|z-2| < 3$
$$\frac{1}{z-5} = \frac{1}{z-2 - 3} = -\frac{1}{3}\frac{1}{1-\frac{z-2}{3}} = -\frac{1}{3}\sum_{n=0}^{\infty} \frac{(z-2)^n}{3^n} = \sum_{n=0}^{\infty} \frac{(z-2)^n}{3^{n+1}}$$
For $|z-2| > 3$
$$\frac{1}{z-5} = \frac{1}{z-2 - 3} = \frac{1}{z-2}\frac{1}{1-\frac{3}{z-2}} = \frac{1}{z-2}\sum_{n=0}^{\infty} \frac{3^n}{(z-2)^n} = \sum_{n=0}^{\infty} \frac{3^n}{(z-2)^{n+1}}$$
