How do I find the residue of a function $\sin(1/z)/(1-z)$ at $z=0$ ?
$z=1$ is a simple pole and residue of this function is $-sin1$.
When at I write Laurent series of $\sin(1/z),$ $z=0$ appears to be a simple pole too, but something goes wrong, and I don't come to the right solution.
$\sin(1/z)$ has an essential singularity at $z=0$. It's Laurent series around $z=0$ is $$ \sin\frac1z=\frac1z-\frac{1}{3!\,z^3}+\frac{1}{5!\,z^5}-\frac{1}{7!\,z^7}+\dots,\quad |z|>0. $$ $1/(1-z)$ is analytic on $\{|z|<1\}$. It's power series expansion around $z=0$ is $$ \frac1{1-z}=1+z+z^2+z^3+\dots,\quad |z|<1. $$ You find the Laurent series of $\sin\dfrac1z\,\dfrac{1}{1-z}$ multiplying the two expressions. To find the residue you need the coefficient of $1/z$. This is obtained multiplying terms whose product gives the power $1/z$: $$ \operatorname{Res}\Bigl(\sin\frac1z\frac1{1-z};0\Bigr)=1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\dots=\sin1. $$