If $\sum_{-\infty}^{\infty}a_n (z+1)^n$ is the Laurent series expansion of $f(z)=\sin{(\frac{z}{z+1})}$, then find the value of $a_{-2}$.
My work:
So we are asked to find the laurent series expansion of $f(z)$ at $z=-1$. Let $u=z+1$ so $z=u-1$, hence $$\begin{align} &\sin\left(\frac{z}{z+1}\right)\\&=\sin\left(\frac{u-1}{u}\right)\\&=\sin\left(1-\frac{1}{u}\right)\\&=\sin(1)\cos\left(\frac{1}{u}\right)-\cos(1)\sin\left(\frac{1}{u}\right)\\ &=\sin(1)\left\{1-\frac{1}{2!u^2}+\frac{1}{4!u^4}-\cdots\right\}+\cos(1)\left\{\frac{1}{u}-\frac{1}{3!u^3}+\cdots\right\}\\ &=\sin(1)\left\{1-\frac{1}{2!(z+1)^2}+\frac{1}{4!(z+1)^4}-\cdots\right\}+\cos(1)\left\{\frac{1}{(z+1)}-\frac{1}{3!(z+1)^3}+\cdots\right\}\end{align}$$
So $a_{-2}=-\frac{\sin(1)}{2}$.
Is my solution correct? Any help is appreciated. Thanks.