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Question is to find coefficient of $(z-\pi)^2$ in taylor series expansion about $\pi$ of
$$f(z)=\frac{\sin z}{z-\pi},z\neq\pi$$ $$=-1,z=\pi$$
now, coefficient of $(z-\pi)^2$ should be $\frac{f^{''}(\pi)}{2}$. But $f^{''}(\pi)$ is not defined. moreover if a function is not defined at a point, then it is supposed to have Laurent Series expansion about that point, then why is question still asking taylor series expansion of f(z)?

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  • $\begingroup$ It is defined. The value at $z = \pi$ is explicitly given, and the resulting function is holomorphic. $\endgroup$ – Daniel Fischer Mar 28 '14 at 18:44
  • $\begingroup$ so, $f^{'}(z)=0$ then? $\endgroup$ – ketan Mar 28 '14 at 18:46
  • $\begingroup$ $f'(\pi)$ happens to be $0$, if that's what you mean. $\endgroup$ – Daniel Fischer Mar 28 '14 at 18:49
  • $\begingroup$ that way, won't required coefficient be 0 too $\endgroup$ – ketan Mar 28 '14 at 18:50
  • $\begingroup$ No, the required coefficient is not $0$. $\endgroup$ – Daniel Fischer Mar 28 '14 at 18:53
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Instead of using the definition of the Taylor series with the derivatives, if possible, it is usually much easier to work with known series expansions. Here, a trigonometric identity gives us

$$\frac{\sin z}{z-\pi} = \frac{\sin \bigl(\pi + (z-\pi)\bigr)}{z-\pi} = - \frac{\sin (z-\pi)}{z-\pi},$$

and the Taylor expansion about $\pi$ is readily derived.

Moreover if a function is not defined at a point, then it is supposed to have Laurent series expansion about that point, then why is the question still asking Taylor series expansion

When a function $f$ is not defined at a point $z_0$, but is defined and holomorphic in a punctured disk $D_r(z_0)\setminus \{z_0\}$ around that point, then $z_0$ is an isolated singularity of $f$. Then $f$ has a Laurent series expansion

$$f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n = \underbrace{\sum_{n=-\infty}^{-1} a_n(z-z_0)^n}_{\text{principal part}} + \underbrace{\sum_{n=0}^\infty a_n(z-z_0)^n}_{\text{secondary part}}$$

valid in the punctured disk $D_r(z_0)\setminus \{z_0\}$. If it happens that the principal part of the Laurent series vanishes ($a_n = 0$ for all $n < 0$), then $z_0$ is a removable singularity of $f$, and the secondary part of the Laurent series is the Taylor series of the function obtained by removing the removable singularity (by assigning the value $a_0$ in $z_0$). The converse of course also holds, if $z_0$ is a removable singularity of $f$, then the principal part of the Lauren series vanishes, and the Laurent series is the Taylor series of the extended function.

This is precisely the situation we have here. $\sin z$ has a zero in $z_0 = \pi$, hence the function

$$\frac{\sin z}{z-\pi}$$

has a removable singularity in $z_0 = \pi$, and the value of the holomorphic extension in $\pi$ is $-1$, so the given function

$$f(z) = \begin{cases} \dfrac{\sin z}{z-\pi} &, z \neq \pi\\ \;\; -1 &, z = \pi\end{cases}$$

is in fact an entire function, and its Laurent series (about any point) coincides with its Taylor series (about the same point).

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  • $\begingroup$ so i just somehow need the positive powers of $(z-z_o)$ whenever f(z) is explicitly defined at $z_o$ as in this case.because the traditional definition of taylor expansion is of no use here even though f(z) has taylor expansion $\endgroup$ – ketan Mar 28 '14 at 19:19
  • $\begingroup$ Well, if $f(z)$ is defined in the right way at $z_0$. After all, the singularity can be a pole or essential, then defining $f$ at $z_0$ doesn't give you a holomorphic function. You can use the traditional definition of the Taylor series, i.e. explicitly compute the derivatives (by sum, product, quotient and chain rule for $z \neq z_0$, and by the limit of difference quotients [or limit of the derivatives] for $z_0$), that is just usually a lot harder than reusing a known expansion and just dividing that through when that is possible. $\endgroup$ – Daniel Fischer Mar 28 '14 at 19:27

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