# What is the coefficient of $(z-\pi)^2$ in Taylor series expansion of $\sin (z)/ (z-\pi)$

I want to determine the coefficient of $(z- \pi)^2$ in Taylor series expansion of $f(z)=\sin (z)/ (z-\pi)$ if $z \neq \pi$, $-1$ if $z=\pi$ around $\pi$.

How can this be done? I don't know how to do apply Taylor theorem here.

• @MattSamuel I had not seen your comment. I mainly meant to clean up the post.// Ramdev: please provide some details about what you know about the problem.
– quid
Jan 27 '15 at 15:09
• find the coefficient of (z−π)^2. Jan 27 '15 at 15:09
• i dont know how to do apply taylor theorem here. Jan 27 '15 at 15:12

It will be the same as the coefficient of $(z-\pi)^3$ in the Taylor series expansion of $f(z) = \sin(z)$ around $z = \pi$. This is just $f^{\prime\prime\prime}(\pi)/3!$.

Here is a start

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

Now use the identity

$$\sin( a+b )= \sin s \cos b + \sin b \cos a$$

and the power series of $\sin(t)$ to finish the problem.

• thanks. but can we solve this problem in any another approach Jan 27 '15 at 15:20
• i solve it and got the answer 1/6 Jan 27 '15 at 15:22
• @Ramdev: I gave you this approach because it depends on using the Laurent series of $f(z)$ which what they usually teach! Jan 27 '15 at 15:22
• @Ramdev: Yes, it is correct! Good job! Jan 27 '15 at 15:23