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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.

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  • $\begingroup$ @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. $\endgroup$
    – quid
    Jan 27 '15 at 15:09
  • $\begingroup$ find the coefficient of (z−π)^2. $\endgroup$
    – Ramdev
    Jan 27 '15 at 15:09
  • $\begingroup$ i dont know how to do apply taylor theorem here. $\endgroup$
    – Ramdev
    Jan 27 '15 at 15:12
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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 $f^{\prime\prime\prime}(\pi)/3!$.

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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.

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

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