I'm learning complex analysis, specifically Laurent series and isolated singularities, and need help to understand the solution to the following exercise:
Find and determine the nature of the isolated singularity of the function
$$h(z) = \frac{\sin(\pi z)}{z - 1}.$$
Here's the solution:
The function $h(z)$ has one isolated singularity at the point $z_0 = 1$. Also, the function $g(z) = \sin(\pi z)$ is analytic in $\mathbb{C}$. Therefore, by Taylor's theorem, we can write $g(z)$ as a power series centered at one of the form
$$g(z) = \sum_{n = 0}^{\infty} \frac{g^{(n)}(1)}{n!}(z - 1)^n \quad\quad (1)$$
Now for the part that I don't understand. If follows that
$$h(z) = \frac{g(1)}{z - 1} + \sum_{n = 0}^{\infty} \frac{g^{(n)}(1)}{n!}(z - 1)^n \quad\quad (2)$$
Since $g(1) = \sin(\pi) = 0$, then
$$h(z) = \sum_{n = 0}^{\infty} \frac{g^{(n)}(1)}{n!}(z - 1)^n \quad\quad (3)$$
and $z_0 = 1$ is a removable singularity.
I don't understand how we go from $(1)$ to $(2)$. Since $g(1) = 0$, step $(2)$ to $(3)$ is obvious. Moreover, since the series in $(3)$ has only positive exponents the point $z_0 = 1$ is a removable singularity. It is the second step that causes me problem. I'm also interested to know if there are other methods to determine the nature of the isolated singualarity of $h(z)$.
As you note, $g$ is holomorphic on a disc centred at $1$ (in fact, everywhere).
So we know that $h$ is holomorphic everywhere except at $1$.
Moreover, $h$ is bounded on all sufficiently small discs centred at $1$, by the existence of (a variant of) the famous limit $$\lim_{z \to 1}\frac{\sin(\pi z)}{z-1} = -\pi$$ As can be verified by the Taylor expansion of $\sin$. This is not, as you claim, a consequence of $g(1)$ being $0$. Indeed, if we had divided by $(z-1)^{2}$, then the singularity would not be removable.
Now, since $h$ is holomorphic and bounded near $1$, we can define $h(1)=\pi$, and the resulting function is holomorphic. Note that before you make this step, $h$ is not even defined at $1$!
Now we can apply Taylor's Theorem directly to get the power series expansion $$h(z)=\sum_{n=1}^{\infty}\frac{g^{(n)}(1)}{n!}(z-1)^{n-1}$$
The derivation you give is meaningless. $(2)$ doesn't make sense precisely unless $z \ne 1$ (and you need to change your index of summation!), the idea is that you've taken the constant term out of the integral and divided by $z-1$. the correct way to write this is to say there is no constant term, and then either make a remark about removable singularities, or claim that the power series is obviously given by the divided one.