"For what complex numbers $c$ does the function $f(z)=\frac{1}{z-c}$ have a Laurent expansion on the annulus $A=\{z : 1<|z|<2\}$? Determine this expansion explicitly."
Since $f(z)$ has a singularity at $z=c$, $\ f(z)$ is holomorphic on the domains: $|z|<c$ and $c<|z|<\infty$, which implies $c=1$ and $c=2$, right?
So would the solution be: $\frac{1}{z-1}= \sum_{n=0}^{\infty}\frac{1}{z^{n+1}} \ (1<|z|)$,$\ $ $\frac{1}{z-2}= \sum_{n=0}^{\infty}\frac{-z^n}{2^{n+1}} \ (|z|<2)$?
Thank you for any help!
Since $f$ is analytic on $A$, you must have either $|c| \le 1$ or $|c| \ge 2$.
In the first case, if $z \in A$, then $|z| > |c|$. Write $f(z) = {1 \over z} {1 \over 1-{c \over z}}$.
In the second case, if $z \in A$, then $|z| < |c|$. Write $f(z) = -{1 \over c} {1 \over 1-{z \over c}}$.