taylor series expansion about undefined point

Question

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)?

Answer 1

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).