Problem:
Obtain the following Laurent expansions. State the first four nonzero terms. State explicitly the $n$th term in the series, and state the largest possible annular domain in which your series is a valid representation of the function.
- $f(z) =\sin\left(1+\frac{1}{z-1}\right)$ expanded in powers of $z-1$
Attempt at solution:
I had already solved everything (and checked my solutions with those in the back of the book) except the very last part concerning the annular domain. Now, I know that if the function is analytic in the annular domain $D = r_1 < |z-z_0| < r_2$, then the function can be represented by a Laurent series in powers of $z-z_0$.
Thus, for this particular case, my $z_0=1$. Now, I need to find where $f(z)$ is not analytic. I know that $\sin(z)$ is entire (analytic everywhere) and I know that $1+\frac{1}{z-1}$ is analytic everywhere except at the isolated singularity when $z=1$. Thus, by composing these functions, $f(z)$ is analytic everywhere except at $z=1$. Thus, it is analytic in the domain $0<|z-1|<\infty$. Thus, it seems I would be able to conclude that the Laurent expansion is valid in $0<|z-1|<\infty$, but according to my textbook, the solution is $|z-1|>1$. I've been thinking about this all day but I can't seem to convince myself of the solution in the textbook so any help is appreciated.
Note that $$f(z)=\sin(1)\cos\left(\frac1{z-1}\right)+\cos(1)\sin\left(\frac1{z-1}\right),$$ and recall the well-known expansions, valid for every $u$ in $\mathbb C$, $$ \cos(u)=\sum_{n\geqslant0}(-1)^n\frac{u^{2n}}{(2n)!}, \qquad \sin(u)=\sum_{n\geqslant0}(-1)^n\frac{u^{2n+1}}{(2n+1)!}. $$ Thus, on the annular $\mathbb C\setminus\{1\}$, $$ f(z)=\sum_{n\geqslant0}a_n\cdot\left(\frac1{z-1}\right)^n, \qquad a_n=\text{____}. $$