I'm uncertain about poles and removable singularities. The function $g(z) = \frac{2z}{(1-z^2)^2} = 2z \cdot \frac{1}{(1-z)^2} \cdot \frac{1}{(1+z)^2}$ has singular points at $z_1 = 1, z_2 = -1$.
Applying
$$
\operatorname{Res}(f, c)=\frac{1}{(n-1) !} \lim _{z \rightarrow c} \frac{d^{n-1}}{d z^{n-1}}\left((z-c)^n f(z)\right)
$$
for a poles of order $n=2$,
$$Res(f, 1) =\frac{1}{1 !} \lim _{z \rightarrow 1} \frac{d^{1}}{d z^{1}}\left((z-1)^2 \cdot 2z \cdot \frac{1}{(1-z)^2} \cdot \frac{1}{(1+z)^2}\right)$$
$$Res(f, 1) = \lim _{z \rightarrow 1} \frac{d}{d z}\left(2z \cdot \frac{1}{(1+z)^2}\right) = \lim _{z \rightarrow 1} 2 \cdot \frac{1-z}{(1+z)^3} = 0$$
So, this means $z=1$ is not actually a pole, but a removable singularity? How would I go about finding the value I would use for g(1)?
Best Answer
As is the case here it's possible for a function to have a residue of $0$ at a pole of order $\geq 2$. Indeed, the Laurent series of $g(z)$ about $z = 1$ is $$\frac{1}{2} \frac{1}{(z - 1)^2} + \boxed{0} \cdot \frac{1}{z - 1} + R(z) ,$$ where $R(z) \in O(1)$. The boxed coefficient is the residue.