Is zero a singular point of this function?

Question

$$f(z)=\frac{z^3}{z+z^5}$$I thought that this function has 5 singular points. But my friend is convinced it only has four because if you write is as$$f(z)=\frac{z^2}{1+z^4}$$ then it is defined at f(0). This is really confusing me. Looking at the original function, it seems to me like 0 should be a (removable) singular point. I mean if you plug z=0 into it, it doesn't looke like f(z) is analytic to me...

Answer 1

You are correct - the function is defined everywhere except at the 4 roots of -1, and at $z=0$. I think the term removable singularity that you used is correct - it is possible to define a function such that: $$f(z) = \begin{cases} \frac{z^3}{z+z^5} & z \neq 0, \\ \\ 0 & z = 0. \end{cases} $$