Find The Laurent Series of $\frac{z}{z-1} \sin{z}$ around $z=1$ We will start by finding the expansion of $\sin{z}$ around $1$.
$$\sin(z) = \frac{1}{2i} \left(e^{iz} - e^{-iz}\right) = \frac{1}{2i} \left(e^{i(z-1+1)} - e^{-i(z-1+1)}\right) = \frac{1}{2i} \left( e^{i(z-1) + i}- e^{-i(z-1)-i} \right)$$
Now we note that $e^{i(z-1)} = \sum_{n=0}^\infty(z-1)^n$ and $e^{-i(z-1)} = \sum_{n=0}^\infty(-1)^n(z-1)^n$. Hence
$$\sin(z) = \sum_{n=0}^\infty \left( \frac{e^i + e^{-i}(-1)^{n+1}}{2i} \right)(z-1)^n$$
From this, we see that
$$z\frac{\sin{z}}{z-1} = z\sum_{n=0}^\infty \left( \frac{e^i + e^{-i}(-1)^{n+1}}{2i} \right)(z-1)^{n-1} = z\sum_{n=-1}^\infty \left( \frac{e^i + e^{-i}(-1)^{n}}{2i} \right)(z-1)^{n}$$
Now we rewrite $z$ as $(z-1)+1$ and obtain:
$$z\frac{\sin{z}}{z-1} = \left[(z-1)+1\right]\sum_{n=-1}^\infty \left( \frac{e^i + e^{-i}(-1)^{n}}{2i} \right)(z-1)^{n}$$
After some manipulation, we get
$$z \frac{\sin{z}}{z-1} = \sum_{n=0}^\infty \frac{e^i}{i}(z-1)^n + \frac{e^i-e^{-i}}{2i} \frac{1}{z-1}$$
Is my method correct? If needed, I can provide the derivation that I skipped. Please let me know if there is a better/more effective method.

My substitution of $e^{i(z-1)} = \sum_{n=0}^\infty(z-1)^n$ and $e^{-i(z-1)} = \sum_{n=0}^\infty(-1)^n(z-1)^n$. Was incorrect, Instead it should be $e^{i(z-1)} = \sum_{n=0}^\infty i^n(z-1)^n$ and $e^{-i(z-1)} = \sum_{n=0}^\infty(-i)^n(z-1)^n$. Putting the correct substitutions into $\sin{z}$ gives us
$$\sin{z}=\sum_{n=0}^\infty \frac{i^n(e^i + e^{-i}(-1)^{n+1})}{2i}(z-1)^n$$
and so
$$\frac{\sin{z}}{z-1} = \sum_{n=0}^\infty \frac{i^n(e^i + e^{-i}(-1)^{n+1})}{2i}(z-1)^{n-1} = \sum_{n=-1}^\infty \frac{i^n (e^i + e^{-i}(-1)^n)}{2}(z-1)^n$$
thus
$$z \frac{\sin{z}}{z-1} = [(z-1)+1]\sum_{n=-1}^\infty \frac{i^n (e^i + e^{-i}(-1)^n)}{2}(z-1)^n$$
After expanding and then simplifying
$$z \frac{\sin{z}}{z-1} = \sum_{n=0}^\infty \frac{1}{2} \left( i^{n-1}(e^i - e^{-i}(-1)^{n-1}) + i^n(e^i+e^{-i}(-1)^n)\right) (z-1)^n$$
From here, I do not know how to proceed. There does not seem to be any easy way to simplify.
 A: An  alternative approach: $\frac z {z-1} \sin z=\frac {(z-1)+1} {z-1} [\sin (z-1)\cos 1+\cos (z-1)\sin 1]$. Use the series expansion of $\sin (z-1)$ and $\cos (z-1)$
A: Hint: Since the series expansion of the exponential function is $e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}$ we have
\begin{align*}
e^{i(z-1)}=\sum_{n=0}^{\infty}i^n\frac{(z-1)^n}{n!}
\end{align*}

We obtain using (1)
\begin{align*}
\color{blue}{\sin(z)}&=\frac{1}{2i}\left(e^{iz}-e^{-iz}\right)\\
&=\frac{1}{2i}\left(e^{i(z-1+1)}-e^{-i(z-1+1)}\right)\\
&=\frac{1}{2i}\left(e^i\sum_{n=0}^{\infty}i^n\frac{(z-1)^n}{n!}-e^{-i}\sum_{n=0}^{\infty}(-i)^n\frac{(z-1)^n}{n!}\right)\tag{$\to$ (1)}\\
&=\frac{1}{2i}\left(\sum_{n=0}^{\infty}\left(i^ne^i-(-i)^ne^{-i}\right)\frac{(z-1)^n}{n!}\right)\\
&=\frac{1}{2i}\left(\sum_{n=0}^{\infty}\left(i^{(2n+1)}e^i-(-i)^{(2n+1)}e^{-i}\right)\frac{(z-1)^{2n+1}}{(2n+1)!}\right)\\
&\qquad+\frac{1}{2i}\left(\sum_{n=0}^{\infty}\left(i^{(2n)}e^i-(-i)^{(2n)}e^{-i}\right)\frac{(z-1)^{2n}}{(2n)!}\right)\\
&=\sum_{n=0}^{\infty}(-1)^n\left(\frac{e^i+e^{-i}}{2}\right)\frac{(z-1)^{2n+1}}{(2n+1)!}\\
&\qquad+\sum_{n=0}^{\infty}(-1)^n\left(\frac{e^i-e^{-i}}{2i}\right)\frac{(z-1)^{2n}}{(2n)!}\\
&\,\,\color{blue}{=\sin(z-1)\cos(1)+\cos(z-1)\sin(1)}
\end{align*}
in accordance with the expansion of $\sin(z)$ in @geetha290km's answer.

