Find the first 6 terms of Laurent Series of $f(z)=\sin(\frac{1}{1-z})$
My attempt:
We need the Laurent expansion of the function $f(z)=\sin(\frac{1}{1-z})$
Note: $\sin(\frac{1}{1-z})=\sin(-\frac{1}{z-1})=-\sin(\frac{1}{z-1})$ this function isn't analytic in z=1.
The representation of the $\sin$ function around of $z=1$ is:
$\sin(z-1)=\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}(z-1)^{2n+1}$
Then:
$-\sin(\frac{1}{z-1})=\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}(\frac{1}{z-1})^{2n+1}=-\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}({z-1})^{-2n-1}$
Then the first 6 terms are the first 6 terms of the sum $\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}({z-1})^{-2n-1}$ is correct?