Isolated singularities sort Can anyone help me out with finding the nature of the singularities of the following function:
$$g(z)=\frac{\cos z-1}{z^5}$$ without using Taylor expansions? 

You are trying to figure out the nature of the singularity $z_0=0$. Write $p(z)=\cos(z)-1$ and $q(z)=z^5$. Then $g(z)=\frac{p(z)}{q(z)}$. Notice that $z_0$ is a zero of order $2$ for $p$ and a zero of order $5$ for $q$. Then you must have seen a theorem that states that $z_0$ is then a pole of order $5-2=3$. 
Otherwise, write $g(z)=\frac{1}{z^3}f(z)$ with $f(z)=\frac{\cos(z)-1}{z^2}$. Use Hospital's rule to compute that $f(z_0)=-\frac{1}{2}$ and notice that $f$ is holomorphic on $\mathbb{C}$. Then take its Taylor series and multiply by $\frac{1}{z^3}$ to find the Laurent series of $g$. I let you compute its residue at $z_0$.
