Is $z=0$ pole of $\frac{e^z - 1}{z(z-1)(z + i)^2}$ or not?

Question

In the above problem, is z=0 should be a pole.

Because Wolfram is not showing it as a pole.

Please tell me if it is not a pole and why?

Wolfram URL to Check: http://www.wolframalpha.com/input/?i=residues+of+(e%5Ez-1)%2F(z(z%2Bi)%5E2(z-1))+with+%7Cz%7C%3C2

Answer 1

It is not: the numerator and denominator both vanish with multiplicity 1 so the singularity is removable.

Answer 2

If you look at the series centered at $z=0$ you get

$$\frac{e^z-1}{z} = \frac{(1+z+z^2/2+\cdots) -1}{z} = \frac{z+z^2/2+\cdots}{z} = 1+z/2 + \cdots,$$

which is analytic at $z=0$. That is, there's a removable singularity at $z=0.$