Question: Co-efficient of $(z-\pi)^2$ Taylor Series Expansion of $f$ around $\pi$ is

$1.1/2$

$2. -1/2$

$3.1/6$

$4.-1/6$

where $f(z)={\sin z\over z-\pi};z\ne \pi$ and 
$-1$ at $z=\pi$

Could any one tell me how to solve these kind of problem?

Answer: 
Another approach: let $z=w+\pi$. Then

$$f(w+\pi) = \frac{\sin{(w+\pi)}}{w} = -\frac{\sin{w}}{w}$$

Now expand about $w=0$ as usual.
Expand $\sin z$ as a Taylor series about the point $z=\pi$.