Laurent series, am I correct in this reasoning? A problem in my book asks:
In the Laurent series for $\displaystyle f(z) = \frac{1}{(z-4)}$ centered at $z=1$, what is the coefficient of $(z-1)^{-2}$?
The book's solution gives
$$\frac{1}{4-z} = \frac{1}{z-1-3} = \frac{\frac{1}{z-1}}{(1-\frac{3}{(z-1)})} = \frac{1}{z-1} \sum_{n=0}^\infty (\frac{3}{z-1})^n,$$
so the cofficient will be $3$.
However, I don't think this is correct (I don't know complex variables that well, so I'm hesitant to say the book is wrong). First, $f(z)$ is analytic at $z=1$, so shouldn't we get a unique power series expansion about $1$, and hence no negative coefficients?
Second, I'm not sure their expansion of $\displaystyle 1/(1-\frac{3}{(z-1)})$ is valid near $1$, since $\displaystyle |\frac{3}{(z-1)}|$ is not less than $1$ for $z$ close to $1$.
Can anyone confirm my reasoning, or explain why it's wrong?
The way I did this problem was:
$$\frac{1}{z-4} = \frac{1}{(z-1)-3} = \frac{-1}{3-(z-1)} = \frac{-1/3}{1-(z-1)/3}$$
and so we get
$$f(z) = -\frac{1}{3} \sum_{n=0}^\infty \left(\frac{z-1}{3}\right)^n$$
 A: It said "centered at 1"; it didn't say "near 1".  There is a Laurent series, centered at 1, valid in $|z-1|<3$.  But there is a separate Laurent series, centered at 1, valid in $|z-1|>3$.  The first of these, is, indeed, a Taylor series.  But the second of these is the one in the book's solution.
A: You are right and the book is wrong, or worse than that, is deliberately setting a trap to the student. Here is why.   
If you want to study a meromorphic function $f$ near  $z=z_0$, you often  use its  canonical development $f(z)= \sum a_n(z-z_0)^n$ in a Laurent series at $z_0$, with only finitely many negative $n$ for which $a_n\neq 0$. This will tell you  the most important things about the germ of $f$ , namely whether the singularity at $z=z_0$ is removable ($\iff a_n=0 \text { for n}\lt 0$) and else the value of its residue at $z_0$, namely $a_{-1}$ .
From the "solution" in your text one would conclude (as you correctly did ) that $f$ has an essential singularity and that the residue is $1$, which are both ridiculous.  
Now it is true that if you fix an annulus $A=Ann (z_0; r,R)$ around $z_0$ and consider a holomorphic finction $f\in \mathcal O(A)$ on $A$, you will obtain from these data a Laurent series $L(f;A)=\sum \limits_ {n=-\infty}^{\infty} a_n(z-z_0)^n$, where equality is in the sense of compact (and even normal) convergence on $A$.
But then you must say in advance what the annulus $A$ is. The "solution" in the book would only be correct if you took $A=Ann (1; r,R)$ with $3\lt r \lt R$ which is artificial and most importantly should be stated explicitly in the question.
By the way, I heartily congratulate you on your critical reading of the book and the accurateness of your remarks and objections : this is exactly the right way to study mathematics.
