I'm having a little trouble, calculating the residues of $ f(z)=\frac{1}{z} \sin(\frac{1}{1-z} ) $ at $z=0$ and $z=1$.
As for the residue at $z=0$, it can easily be calculated since $0$ is a simple pole of $f$, we have : $Res(f,1)=\lim_{z\to 0} zf(z)=\sin(1)$
However, it seems to me that there are two ways to calculate the residue of $f$ at $1$ :
If I calculate the Laurent Series of f, I get one answer : With $u=z-1$
$$ f(z)= \frac{1}{1+u}\sin(\frac{-1}{u})= (-1) \times (\sum_{n=0}^{\infty }{(-1)^n u^n}) \times (\sum_{n=0}^{\infty }{\frac{(-1)^n}{(2n+1)!}u^{-n}})$$
Thus, the residue is equal to the $u^{-1}$ coefficient : $$ (-1) \times \sum_{n=0}^{\infty }{(-1)^n \times \frac{(-1)^{n+1}}{(2n+3)!} }=\sum_{n=0}^{\infty }{ \frac{1}{(2n+3)!} } = \sum_{n=0}^{\infty }{ \frac{1}{(2n+1)!} } - 1 = \sin(-1)-1 \\= -1-\sin(1)$$
Thus, by calculating the Laurent series, I get : $Res(f,1) = -1-\sin(1)$
However, through the residue theorem, I get another answer :
If we consider the circle of radius $R>1$, centered in $1$, then through the residue theorem : $$ \int_{C_ R}{f(z)\,dz} = 2i\pi(Res(f,0) + Res(f,1)) $$
Furthermore, since $ \lim_{|z|\to\infty} zf(z) =0 $ , through Jordan's lemma, we get : $$ Res(f,0) + Res(f,1) = \frac{1}{2i\pi} \times \lim_{R\to\infty} \int_{C_ R}{f(z)\,dz}=0 $$
Therefore : $ Res(f,1)=-Res(f,0)=-\sin(1) $
I seem to get two different answers, and I don't see where I made a mistake. Any help would be much appreciated, thanks.
As I said in the comment the error in your computation is in the expansion of the sine function. With a correct expansion of the sine we get:
$$ \begin{align*} \frac1{z}\sin \left(\frac1{1-z}\right)&=-\frac1{1+(z-1)}\sin \left(\frac1{z-1}\right)\\ &=-\sum_{k\geqslant 0}(-1)^k(z-1)^k \sum_{j\geqslant 0}(-1)^j\frac{(z-1)^{-2j-1}}{(2j+1)!} \end{align*} $$
Therefore the coefficient for $(z-1)^{-1}$ is given by
$$ -\sum_{k\geqslant 0}(-1)^{2k}\frac{(-1)^k}{(2k+1)!}=-\sum_{k\geqslant 0}(-1)^{k}\frac{1^{2k+1}}{(2k+1)!}=-\sin (1) $$ ∎