I beg you help me to solve this.
I used Taylor expansion of $\sin x$ and then placed $\dfrac{z}{1-z}$ as $x$ but the answer is different.
This is the answer of textbook AND this is my answer
What am I doing wrong, is it the same thing ?
pls help any hint would be priceless
What you have produced isn't a Laurent series. A Laurent series around $z=1$ needs to be in the form $\sum_{n=-\infty}^\infty a_n (z-1)^n$.
To obtain what you need we need to write $$ \sin\big(\frac{z}{z-1}\big) = \sin\big(1+\frac{1}{z-1}) = \sin 1 \cos\frac{1}{z-1} + \cos 1 \sin\frac{1}{z-1}$$ Only then we use the Taylor series for $\sin x$ and $\cos x$, obtaining $$ \sin\big(\frac{z}{z-1}\big) = \sum_{n=0}^\infty \frac{(-1)^n \sin 1}{(2n)!(z-1)^{2n}} + \sum_{n=0}^\infty \frac{(-1)^n \cos 1}{(2n+1)!(z-1)^{2n+1}}$$ Then we use the formulae $$ (-1)^n\sin 1 = \sin(1+n\pi), \qquad (-1)^n\cos 1 = \sin\big(1+ \frac{(2n+1)\pi}{2}\big)$$ to get $$ \sin\big(\frac{z}{z-1}\big) = \sum_{n=0}^\infty \frac{\sin \big(1+\frac{2n}{2}\pi\big)}{(2n)!(z-1)^{2n}} + \sum_{n=0}^\infty \frac{\sin \big(1+\frac{2n+1}{2}\pi\big)}{(2n+1)!(z-1)^{2n+1}} = \sum_{k=0}^\infty \frac{\sin \big(1+\frac{k}{2}\pi\big)}{k!(z-1)^{k}}$$