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
$f(z) = -\frac{1}{3} \sum_{n=0}^\infty \left(\frac{z-1}{3}\right)^n$