So I see in my textbook that the formula for Laurent series was this:
$$\begin{align}f(z) &= \sum_{n=0}^{\infty} a_n(z - z_0)^n + \sum_{n=1}^{\infty} \frac{b_n}{(z - z_0)^n} \\[1ex] &= a_0 + a(z - z_0) + a_2(z - z_0)^2 + \cdots + \frac{b_1}{(z-z_0)} + \frac{b^2}{(z-z_0)^2} \end{align}$$
So I'm confused when the solution for this Laurent series just looks like a Taylor Series that was compared with another simpler Taylor series. It looks like that's what was done here in this problem:
Expand the function in a Laurent series that converges for $0 < |z| < R$ and determine the precise region of convergence. Show the details of your work. $$\frac{\cos{z}}{z^4}$$
So I know that a Taylor expansion of $\cos{z} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} z^{2n}$ and the region of convergence is the whole complex plane. This means that:
$$\begin{align}f(z) &= \frac{\cos{z}}{z^4} \\[1ex] &= \frac{1}{z^4} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} z^{2n} \\[1ex] &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} z^{2n - 4} \end{align}$$
What is the region of convergence and how do you know? Is this a Laurent series? Can this have been solved by a Taylor series? Can't Taylor series be sufficient here? Why or why not? Can't Taylor series handle the singularity at $z = 0$? I guess I don't know why what I did is a Laurent series? Is it because it has some negative exponent terms?