Classify singularities of $f(z) = \frac{z^5}{1+z+z^2+z^3+z^4}$

Question

I'm trying to classify the singularities of $f(z) = \frac{z^5}{1+z+z^2+z^3+z^4}$ and I was wondering if there was an easier way than going to the Laurent series expansion of the function around each point. I know (or I am pretty sure) that the singularities are at $\{e^\frac{2\pi}{5}, e^\frac{4\pi}{5}, e^\frac{6\pi}{5}, e^\frac{8\pi}{5}, \infty\}$. I know there's a Lemma that says

$z_0$ is a removable singularity of $f \iff \lim_{z\rightarrow z_0}(z-z_0)f(z) = 0$

So using this, I showed that the finite singularities weren't removable, but I didn't really know how to go any further.

Answer 1

Hint:

$$f(z) = \frac{z^5}{1+z+z^2+z^3+z^4} = \frac{z^5}{\frac{z^5-1}{z-1}} = \frac{z^5(z-1)}{z^5-1} =\frac{(z^5-1)(z-1)+(z-1)}{z^5-1}= (z-1)\left(1 + \frac{1}{z^5-1}\right)$$