finding singular points of a complex function

Question

I am trying to find the singular points of the function

$$ f(z)= z^4/(z-z^5) $$

which would have the singular points $0,1,-1,i,-i$.

But I am not sure if I should simplify it to

$$ f(z)= z^3/(1-z^4) $$

which would have the singular points $1,-1$.

Thanks for your help!

Answer 1

$z^{4}=1$ does not imply $z =\pm 1$. So in the second form also the singularities are $\pm 1, \pm i$. $\, 0$ has to be excluded from the domain but the function has a removable singularity at $0$.