Question Find and classify the singularities of $f(z)=\dfrac{e^{-z}\sin(2(z-1)^2)}{(z^2-4)(z-1)^2}$.
My Answer
The denominator has zero's at $-2,1$ and $2$. I found both $-2$ and $2$ to be simple poles as the denominator has a zero of order $1$ at $z=2,-2$ and the numerator does not vanish at either point.
I am unsure how to classify $z=1$. I looked online and found that if the function $g(z)=\frac{\sin(2(z-1)^2)}{z-1}$ is analytic then $z=1$ is a simple pole. I found the Taylor expansion of this function to be, $\frac{\sin(2(z-1)^2)}{z-1}=2(z-1)-\frac{2^3(z-1)^7}{3!}+\frac{2^5(z-1)^9}{5!}-\cdots$ but I dont understand why this means that $g(z)$ is analytic or why this implies that $z=1$ is a simple pole. Am I going about this in the right way or is there a simpler method to classify this singularity?
How I calculated the Taylor expansion
$$\sin(z)= z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$$
$$\sin((z-1)^2) = (z-1)^2 - \frac{(z-1)^6}{3!} + \frac{(z-1)^{10}}{5!} - \cdots$$
$$\sin(2(z-1)^2) = 2(z-1)^2 - \frac{2^3(z-1)^8}{3!} + \frac{2^5(z-1)^{10}}{5!} - \cdots$$
$$\frac{\sin(2(z-1)^2)}{z-1} = 2(z-1) - \frac{2^3(z-1)^7}{3!} + \frac{2^5(z-1)^9}{5!} - \cdots$$
Ignore the part of the function which is analytic at $z=1$, namely $\frac{e^{-z}}{z^2-4}$, and consider the remaining part, $g(z)=\frac{sin(2(z-1)^2)}{(z-1)^2}$. Define $h(z)=sin(2(z-1)^2)$. Notice that $h(1)=0, h'(1)=0$ but $h''(1) \neq 0$. Thus $h(z)$ has a zero of order 2 at z=1. Since the denominator of this function also has a zero of order 2 at z=1, the singularity is removable and its residue is zero.